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Formulario de Cálculo Diferencial e Integral                                                                                                                                                                                                                                                                                                                    Jesús Rubí M.

Formulario de                                                                ( a + b ) ⋅ ( a 2 − ab + b2 ) = a3 + b3                                                        θ         sen        cos            tg           ctg            sec        csc
                                                                                                                                                                                                                                                             Gráfica 4. Las funciones trigonométricas inversas
                                                                                                                                                                                                                                                             arcctg x , arcsec x , arccsc x :                     sen α + sen β
                                                                                                                                                                                                                                                                                                                                               1              1
                                                                                                                                                                                                                                                                                                                                        = 2sen (α + β ) ⋅ cos (α − β )
                                                                             ( a + b ) ⋅ ( a3 − a 2 b + ab2 − b3 ) = a 4 − b4                                             0            0       ∞      ∞          0                                                                                                                             2              2
Cálculo Diferencial
                                                                                                                                                                                       1           1
                                                                                                                                                                                                                                                                                                                                               1              1
                                                                                                                                                                          30          12        3 2 3 2                                                                                                           sen α − sen β         = 2 sen (α − β ) ⋅ cos (α + β )
                                                                                                                                                                                                                                                                     4
                                                                                                                                                                                      3 2 1 3
                                                                             ( a + b ) ⋅ ( a 4 − a3b + a 2 b2 − ab3 + b4 ) = a5 + b5
e Integral    VER.4.3                                                                                                                                                     45     1 2 1 2   1   1    2  2                                                             3
                                                                                                                                                                                                                                                                                                                                               2              2
                                                                             ( a + b ) ⋅ ( a5 − a 4 b + a3b2 − a 2 b3 + ab4 − b5 ) = a 6 − b6
                                                                                                                                                                                                                                                                                                                                               1              1
                                                                                                                                                                          60      3 2 12    3 1 3  2 2 3                                                                                                          cos α + cos β         = 2 cos (α + β ) ⋅ cos (α − β )
Jesús Rubí Miranda (jesusrubim@yahoo.com)                                                                                                                                                                                                                            2                                                                         2              2
                                                                                                                                                                          90      1    0   ∞   0   ∞  1
http://mx.geocities.com/estadisticapapers/                                               ⎛ n                     ⎞                                                                                                                                                                                                cos α − cos β
                                                                                                                                                                                                                                                                                                                                                 1              1
                                                                                                                                                                                                                                                                                                                                        = −2sen (α + β ) ⋅ sen (α − β )
                                                                             ( a + b ) ⋅ ⎜ ∑ ( −1) a n− k b k −1 ⎟ = a n + b n ∀ n ∈
                                                                                                  k +1
                                                                                                                                                             impar                             ⎡ π π⎤                                                                1

http://mx.geocities.com/dicalculus/                                                      ⎝ k =1                  ⎠                                                    y = ∠ sen x          y ∈ ⎢− , ⎥                                                                                                                                            2              2
                                                                                                                                                                                               ⎣ 2 2⎦
                                                                                                                                                                                                                                                                                                                                        sen (α ± β )
                                                                                                                                                                                                                                                                     0
                               VALOR ABSOLUTO                                                ⎛   n
                                                                                                                           ⎞
                                                                             ( a + b ) ⋅ ⎜ ∑ ( −1)                                                                                         y ∈ [ 0, π ]
                                                                                                                k +1
                                                                                                            a n − k b k −1 ⎟ = a n − b n ∀ n ∈               par      y = ∠ cos x                                                                                                                                 tg α ± tg β =
     ⎧a si a ≥ 0                                                                             ⎝ k =1                        ⎠
                                                                                                                                                                                                                                                                 -1
                                                                                                                                                                                                                                                                                                                                        cos α ⋅ cos β
 a =⎨
                                                                                                                                                                                                                                                                                                  arc ctg x

                                                                                                                                                                                                          π π                                                                                     arc sec x

     ⎩− a si a < 0                                                                                                                                                    y = ∠ tg x           y∈ −                                                                                                   arc csc x
                                                                                                        SUMAS Y PRODUCTOS                                                                                  ,                                                                                                                     1
                                                                                                                                                                                                                                                                                                                                   ⎡sen (α − β ) + sen (α + β ) ⎤
                                                                                                                                                                                                                                                                 -2
                                                                                                                                                                                                          2 2                                                      -5                     0                   5
                                                                                                                                                                                                                                                                                                                  sen α ⋅ cos β =
 a = −a                                                                                                               n
                                                                                                                                                                                                                                                                                                                                 2⎣                             ⎦
                                                                             a1 + a2 +               + an = ∑ ak                                                                         1                                                                           IDENTIDADES TRIGONOMÉTRICAS
a ≤ a y −a≤ a                                                                                                                                                         y = ∠ ctg x = ∠ tg                   y ∈ 0, π                                                                                                              1
                                                                                                                  k =1
                                                                                                                                                                                         x                                                                                                                        sen α ⋅ sen β = ⎡cos (α − β ) − cos (α + β ) ⎤
                                                                                                                                                                                                                                                             sen θ + cos2 θ = 1                                                  2⎣                            ⎦
                                                                                                                                                                                                                                                                 2
                                                                              n
 a ≥0y a =0 ⇔ a=0
                                                                             ∑ c = nc                                                                                 y = ∠ sec x = ∠ cos
                                                                                                                                                                                          1
                                                                                                                                                                                             y ∈ [ 0, π ]                                                    1 + ctg 2 θ = csc2 θ                                                1
                                                                             k =1                                                                                                                                                                                                                                 cos α ⋅ cos β = ⎡cos (α − β ) + cos (α + β ) ⎤
                                                                                                                                                                                                                                                                                                                                 2⎣                            ⎦
                                n                     n                                                                                                                                   x
 ab = a b ó                    ∏a             = ∏ ak                          n                   n
                                                                                                                                                                                                   ⎡ π π⎤                                                    tg 2 θ + 1 = sec2 θ
                                                                             ∑ ca           = c ∑ ak                                                                                      1
                                      k
                               k =1               k =1                              k                                                                                 y = ∠ csc x = ∠ sen     y ∈ ⎢− , ⎥                                                                                                                        tg α + tg β
                                          n                   n              k =1                k =1                                                                                     x        ⎣ 2 2⎦                                                    sen ( −θ ) = − sen θ                                 tg α ⋅ tg β =
 a+b ≤ a + b ó                        ∑a              ≤ ∑ ak                  n                           n                    n                                                                                                                                                                                               ctg α + ctg β
                                      k =1
                                                  k
                                                          k =1               ∑(a
                                                                             k =1
                                                                                    k       + bk ) = ∑ ak + ∑ bk
                                                                                                         k =1                 k =1
                                                                                                                                                                     Gráfica 1. Las funciones trigonométricas: sen x ,                                       cos ( −θ ) = cosθ                                               FUNCIONES HIPERBÓLICAS
                                                                                                                                                                     cos x , tg x :
                                      EXPONENTES                              n                                                                                                                                                                              tg ( −θ ) = − tg θ                                            e x − e− x
a p ⋅ a q = a p+q                                                            ∑(a            − ak −1 ) = an − a0                                                                                                                                                                                                   senh x =
                                                                                                                                                                                                                                                             sen (θ + 2π ) = sen θ
                                                                                    k                                                                                      2

                                                                             k =1
                                                                                                                                                                                                                                                                                                                                2
 ap                                                                                                                                                                      1.5
                                                                                                                                                                                                                                                                                                                           e x + e− x
    = a p−q                                                                   n
                                                                                                                          n                                                                                                                                  cos (θ + 2π ) = cosθ                                 cosh x =
 aq                                                                          ∑ ⎡ a + ( k − 1) d ⎤ = 2 ⎡ 2a + ( n − 1) d ⎤
                                                                               ⎣                ⎦     ⎣                 ⎦
                                                                                                                                                                           1
                                                                                                                                                                                                                                                                                                                                2
                                                                             k =1                                                                                                                                                                            tg (θ + 2π ) = tg θ
( a p ) = a pq                                                                                                                                                                                                                                                                                                            senh x e x − e − x
          q                                                                                                                                                              0.5

                                                                                                        n                                                                                                                                                                                                         tgh x =         =
                                                                                                          (a + l )
                                                                                                                =                                                          0                                                                                 sen (θ + π ) = − sen θ                                       cosh x e x + e − x
(a ⋅ b)           = ap ⋅bp
              p
                                                                                                        2                                                               -0.5
                                                                                                                                                                                                                                                             cos (θ + π ) = − cosθ                                            1       e x + e− x
                                                                              n
                                                                                                 1 − r n a − rl                                                                                                                                                                                                   ctgh x =        =
⎛a⎞  ap
          p
                                                                             ∑ ar   k −1
                                                                                              =a        =                                                                 -1
                                                                                                                                                                                                                                                             tg (θ + π ) = tg θ                                            tgh x e x − e − x
⎜ ⎟ = p                                                                      k =1                1− r      1− r
⎝b⎠
                                                                                                                                                                                                                                           sen x
     b                                                                                                                                                                  -1.5
                                                                                                                                                                                                                                                                                                                               1            2
                                                                                                                                                                                                                                                             sen (θ + nπ ) = ( −1) sen θ                          sech x =          =
                                                                                                                                                                                                                                           cos x                                      n

                                                                                      ( n + n)
                                                                              n
                                                                                    1 2
                                                                             ∑k =
                                                                                                                                                                                                                                           tg x

a   p/q
          = a
                  q       p                                                                                                                                               -2
                                                                                                                                                                            -8   -6        -4        -2     0        2        4        6           8                                                                       cosh x e x + e − x
                                                                                    2
                                                                             k =1
                                                                                                                                                                                                                                                             cos (θ + nπ ) = ( −1) cos θ
                                                                                                                                                                                                                                                                                      n
                                                                                                                                                                                                                                                                                                                               1            2
                   LOGARITMOS                                                                                                                                                                                                                                                                                     csch x =          =
                                                                             ∑ k 2 = 6 ( 2n3 + 3n2 + n )
                                                                              n
                                                                                     1                                                                               Gráfica 2. Las funciones trigonométricas csc x ,
log a N = x ⇒ a x = N                                                                                                                                                                                                                                        tg (θ + nπ ) = tg θ                                           senh x e x − e − x
                                                                             k =1                                                                                    sec x , ctg x :
log a MN = log a M + log a N                                                                                                                                                                                                                                                                                      senh :     →
                                                                                                                                                                                                                                                             sen ( nπ ) = 0
                                                                             ∑ k 3 = 4 ( n 4 + 2n3 + n 2 )
                                                                              n
                                                                                     1
      M
                                                                                                                                                                         2.5
                                                                                                                                                                                                                                                                                                                  cosh :     → [1, ∞
          = log a M − log a N                                                                                                                                                                                                                                cos ( nπ ) = ( −1)
                                                                                                                                                                                                                                                                                  n
log a                                                                        k =1                                                                                          2

      N                                                                                                                                                                                                                                                                                                           tgh :    → −1,1
                                                                             ∑ k 4 = 30 ( 6n5 + 15n4 + 10n3 − n )
                                                                              n
                                                                                      1
                                                                                                                                                                                                                                                             tg ( nπ ) = 0
                                                                                                                                                                         1.5

log a N r = r log a N                                                                                                                                                      1
                                                                                                                                                                                                                                                                                                                  ctgh :     − {0} → −∞ , −1 ∪ 1, ∞
                                                                             k =1
                                                                                                                                                                                                                                                                  ⎛ 2n + 1 ⎞
                                                                                                        + ( 2n − 1) = n 2                                                                                                                                                 π ⎟ = ( −1)                                        → 0,1]
                                                                                                                                                                         0.5                                                                                                          n
                          log b N ln N                                       1+ 3 + 5 +                                                                                                                                                                      sen ⎜
log a N =                         =                                                                                                                                                                                                                                                                               sech :
                                                                                                                                                                                                                                                                  ⎝ 2        ⎠
                                                                                                                                                                           0
                           log b a ln a                                                 n
                                                                                                                                                                                                                                                                                                                             − {0} →          − {0}
                                                                             n! = ∏ k
                                                                                                                                                                        -0.5
                                                                                                                                                                                                                                                                                                                  csch :
log10 N = log N y log e N = ln N                                                                                                                                                                                                                                  ⎛ 2n + 1 ⎞
                                                                                    k =1
                                                                                                                                                                          -1
                                                                                                                                                                                                                                                             cos ⎜        π⎟=0
                ALGUNOS PRODUCTOS                                                                                                                                       -1.5                                                                                      ⎝ 2        ⎠                                    Gráfica 5. Las funciones hiperbólicas senh x ,
                                                                             ⎛n⎞         n!                                                                                                                                                csc x

a ⋅ ( c + d ) = ac + ad                                                      ⎜ ⎟=                 , k≤n                                                                   -2                                                               sec x
                                                                                                                                                                                                                                                                ⎛ 2n + 1 ⎞                                        cosh x , tgh x :
                                                                             ⎝ k ⎠ ( n − k )!k !
                                                                                                                                                                                                                                           ctg x
                                                                                                                                                                                                                                                             tg ⎜        π⎟=∞
                                                                                                                                                                                                                                                                ⎝ 2        ⎠
                                                                                                                                                                        -2.5


( a + b) ⋅ ( a − b) = a − b
                                                                                                                                                                            -8   -6        -4        -2     0        2        4        6           8
                                      2           2
                                                                                                                                                                                                                                                                                                                             5
                                                                                           n
                                                                                              ⎛n⎞
                                                                             ( x + y ) = ∑ ⎜ ⎟ xn−k y k                                                                                                                                                                      π⎞
                                                                                      n
                                                                                                                                                                     Gráfica 3. Las funciones trigonométricas inversas                                                   ⎛
( a + b ) ⋅ ( a + b ) = ( a + b ) = a 2 + 2ab + b2
                                                                                                                                                                                                                                                                                                                             4

                                                                                                                                                                                                                                                             sen θ = cos ⎜θ − ⎟
                                 2
                                                                                         k =0 ⎝ k ⎠                                                                  arcsen x , arccos x , arctg x :                                                                     ⎝   2⎠                                              3



( a − b ) ⋅ ( a − b ) = ( a − b ) = a 2 − 2ab + b 2
                                 2                                                                                                                                                                                                                                                                                           2


                                                                             ( x1 + x2 +                + xk )
                                                                                                                  n
                                                                                                                          =∑
                                                                                                                                   n!
                                                                                                                                           x1n1 ⋅ x2 2                                                                                                                  ⎛   π⎞
                                                                                                                                                                                                                                                             cosθ = sen ⎜θ + ⎟
                                                                                                                                                   n     nk
                                                                                                                                                         x                 4


( x + b ) ⋅ ( x + d ) = x 2 + ( b + d ) x + bd
                                                                                                                                                                                                                                                                                                                             1
                                                                                                                                                         k
                                                                                                                             n1 !n2 ! nk !
                                                                                                                                                                           3
                                                                                                                                                                                                                                                                        ⎝   2⎠                                               0


( ax + b ) ⋅ ( cx + d ) = acx 2 + ( ad + bc ) x + bd                                         CONSTANTES                                                                                                                                                      sen (α ± β ) = sen α cos β ± cos α sen β                        -1




( a + b ) ⋅ ( c + d ) = ac + ad + bc + bd                                    π = 3.14159265359…                                                                            2

                                                                                                                                                                                                                                                             cos (α ± β ) = cos α cos β ∓ sen α sen β
                                                                                                                                                                                                                                                                                                                             -2

                                                                                                                                                                                                                                                                                                                                                                se nh x

                                                                             e = 2.71828182846…
                                                                                                                                                                                                                                                                                                                             -3                                 co sh x


( a + b ) = a3 + 3a 2b + 3ab2 + b3                                                                                                                                                                                                                                          tg α ± tg β
         3                                                                                                                                                                 1                                                                                                                                                                                    tgh x


                                                                                                                                                                                                                                                             tg (α ± β ) =
                                                                                                                                                                                                                                                                                                                             -4
                                                                                                                                                                                                                                                                                                                               -5                       0                 5
                                                                                           TRIGONOMETRÍA
                                                                                                                                                                                                                                                                           1 ∓ tg α tg β
( a − b ) = a3 − 3a 2b + 3ab2 − b3
         3                                                                                                                                                                 0
                                                                                      CO                   1                                                                                                                                                                                                               FUNCIONES HIPERBÓLICAS INV
                                                                             sen θ =             cscθ =                                                                                                                                                      sen 2θ = 2sen θ cosθ
( a + b + c ) = a 2 + b2 + c 2 + 2ab + 2ac + 2bc
               2                                                                     HIP
                                                                                      CA
                                                                                                         sen θ
                                                                                                           1
                                                                                                                                                                          -1
                                                                                                                                                                                                                                   arc sen x
                                                                                                                                                                                                                                   arc cos x
                                                                                                                                                                                                                                   arc tg x                  cos 2θ = cos 2 θ − sen 2 θ                                             (
                                                                                                                                                                                                                                                                                                                  senh −1 x = ln x + x 2 + 1 , ∀x ∈ )
                                                                             cosθ =              secθ =
( a − b ) ⋅ ( a + ab + b ) = a − b                                                                                                                                                                                                                                                                                                  (               )
                                                                                                                                                                          -2

                                                                                                                                                                                                                                                                        2 tg θ
                           2                  2           3       3
                                                                                                         cosθ
                                                                                                                                                                            -3        -2        -1          0            1         2               3
                                                                                     HIP                                                                                                                                                                     tg 2θ =                                              cosh −1 x = ln x ± x 2 − 1 , x ≥ 1
                                                                                    sen θ CO                                                                                                                                                                          1 − tg 2 θ
( a − b ) ⋅ ( a3 + a 2 b + ab2 + b3 ) = a 4 − b4                             tg θ =      =       ctg θ =
                                                                                                          1
                                                                                                                                                                                                                                                                                                                             1 ⎛1+ x ⎞
                                                                                    cosθ CA              tg θ                                                                                                                                                           1                                         tgh −1 x = ln ⎜     ⎟, x < 1
( a − b ) ⋅ ( a 4 + a3b + a 2 b2 + ab3 + b 4 ) = a5 − b5                                                                                                                                                                                                     sen 2 θ = (1 − cos 2θ )                                         2 ⎝ 1− x ⎠
                                                                                                                                                                                                                                                                        2
                      ⎛                       ⎞                              π radianes=180                                                                                                                                                                                                                                   1 ⎛ x +1 ⎞
                                                                                                                                                                                                                                                                                                                  ctgh −1 x = ln ⎜
                           n
                                                                                                                                                                                                                                                                        1
( a − b ) ⋅ ⎜ ∑ a n − k b k −1 ⎟ = a n − b n                          ∀n ∈                                                                                                                                                                                   cos 2 θ = (1 + cos 2θ )                                                    ⎟, x > 1
                                                                                                                                                                                                                                                                                                                              2 ⎝ x −1 ⎠
                      ⎝ k =1                  ⎠                                                                                                                                                                                                                         2
                                                                                                                                                                                                                                                                      1 − cos 2θ                                                 ⎛ 1 ± 1 − x2 ⎞
                                                                                                                       HIP                                                                                                                                   tg 2 θ =                                             sech −1 x = ln ⎜             ⎟, 0 < x ≤ 1
                                                                                                                                                CO                                                                                                                    1 + cos 2θ                                                 ⎜     x       ⎟
                                                                                                                                                                                                                                                                                                                                 ⎝             ⎠
                                                                                                              θ                                                                                                                                                                                                                  ⎛1      x2 + 1 ⎞
                                                                                                                                                                                                                                                                                                                  csch −1 x = ln ⎜ +             ⎟, x ≠ 0
                                                                                                                           CA                                                                                                                                                                                                    ⎜x       x ⎟
                                                                                                                                                                                                                                                                                                                                 ⎝               ⎠
formulas calculo integral y diferencial
formulas calculo integral y diferencial

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formulas calculo integral y diferencial

  • 1. Formulario de Cálculo Diferencial e Integral Jesús Rubí M. Formulario de ( a + b ) ⋅ ( a 2 − ab + b2 ) = a3 + b3 θ sen cos tg ctg sec csc Gráfica 4. Las funciones trigonométricas inversas arcctg x , arcsec x , arccsc x : sen α + sen β 1 1 = 2sen (α + β ) ⋅ cos (α − β ) ( a + b ) ⋅ ( a3 − a 2 b + ab2 − b3 ) = a 4 − b4 0 0 ∞ ∞ 0 2 2 Cálculo Diferencial 1 1 1 1 30 12 3 2 3 2 sen α − sen β = 2 sen (α − β ) ⋅ cos (α + β ) 4 3 2 1 3 ( a + b ) ⋅ ( a 4 − a3b + a 2 b2 − ab3 + b4 ) = a5 + b5 e Integral VER.4.3 45 1 2 1 2 1 1 2 2 3 2 2 ( a + b ) ⋅ ( a5 − a 4 b + a3b2 − a 2 b3 + ab4 − b5 ) = a 6 − b6 1 1 60 3 2 12 3 1 3 2 2 3 cos α + cos β = 2 cos (α + β ) ⋅ cos (α − β ) Jesús Rubí Miranda (jesusrubim@yahoo.com) 2 2 2 90 1 0 ∞ 0 ∞ 1 http://mx.geocities.com/estadisticapapers/ ⎛ n ⎞ cos α − cos β 1 1 = −2sen (α + β ) ⋅ sen (α − β ) ( a + b ) ⋅ ⎜ ∑ ( −1) a n− k b k −1 ⎟ = a n + b n ∀ n ∈ k +1 impar ⎡ π π⎤ 1 http://mx.geocities.com/dicalculus/ ⎝ k =1 ⎠ y = ∠ sen x y ∈ ⎢− , ⎥ 2 2 ⎣ 2 2⎦ sen (α ± β ) 0 VALOR ABSOLUTO ⎛ n ⎞ ( a + b ) ⋅ ⎜ ∑ ( −1) y ∈ [ 0, π ] k +1 a n − k b k −1 ⎟ = a n − b n ∀ n ∈ par y = ∠ cos x tg α ± tg β = ⎧a si a ≥ 0 ⎝ k =1 ⎠ -1 cos α ⋅ cos β a =⎨ arc ctg x π π arc sec x ⎩− a si a < 0 y = ∠ tg x y∈ − arc csc x SUMAS Y PRODUCTOS , 1 ⎡sen (α − β ) + sen (α + β ) ⎤ -2 2 2 -5 0 5 sen α ⋅ cos β = a = −a n 2⎣ ⎦ a1 + a2 + + an = ∑ ak 1 IDENTIDADES TRIGONOMÉTRICAS a ≤ a y −a≤ a y = ∠ ctg x = ∠ tg y ∈ 0, π 1 k =1 x sen α ⋅ sen β = ⎡cos (α − β ) − cos (α + β ) ⎤ sen θ + cos2 θ = 1 2⎣ ⎦ 2 n a ≥0y a =0 ⇔ a=0 ∑ c = nc y = ∠ sec x = ∠ cos 1 y ∈ [ 0, π ] 1 + ctg 2 θ = csc2 θ 1 k =1 cos α ⋅ cos β = ⎡cos (α − β ) + cos (α + β ) ⎤ 2⎣ ⎦ n n x ab = a b ó ∏a = ∏ ak n n ⎡ π π⎤ tg 2 θ + 1 = sec2 θ ∑ ca = c ∑ ak 1 k k =1 k =1 k y = ∠ csc x = ∠ sen y ∈ ⎢− , ⎥ tg α + tg β n n k =1 k =1 x ⎣ 2 2⎦ sen ( −θ ) = − sen θ tg α ⋅ tg β = a+b ≤ a + b ó ∑a ≤ ∑ ak n n n ctg α + ctg β k =1 k k =1 ∑(a k =1 k + bk ) = ∑ ak + ∑ bk k =1 k =1 Gráfica 1. Las funciones trigonométricas: sen x , cos ( −θ ) = cosθ FUNCIONES HIPERBÓLICAS cos x , tg x : EXPONENTES n tg ( −θ ) = − tg θ e x − e− x a p ⋅ a q = a p+q ∑(a − ak −1 ) = an − a0 senh x = sen (θ + 2π ) = sen θ k 2 k =1 2 ap 1.5 e x + e− x = a p−q n n cos (θ + 2π ) = cosθ cosh x = aq ∑ ⎡ a + ( k − 1) d ⎤ = 2 ⎡ 2a + ( n − 1) d ⎤ ⎣ ⎦ ⎣ ⎦ 1 2 k =1 tg (θ + 2π ) = tg θ ( a p ) = a pq senh x e x − e − x q 0.5 n tgh x = = (a + l ) = 0 sen (θ + π ) = − sen θ cosh x e x + e − x (a ⋅ b) = ap ⋅bp p 2 -0.5 cos (θ + π ) = − cosθ 1 e x + e− x n 1 − r n a − rl ctgh x = = ⎛a⎞ ap p ∑ ar k −1 =a = -1 tg (θ + π ) = tg θ tgh x e x − e − x ⎜ ⎟ = p k =1 1− r 1− r ⎝b⎠ sen x b -1.5 1 2 sen (θ + nπ ) = ( −1) sen θ sech x = = cos x n ( n + n) n 1 2 ∑k = tg x a p/q = a q p -2 -8 -6 -4 -2 0 2 4 6 8 cosh x e x + e − x 2 k =1 cos (θ + nπ ) = ( −1) cos θ n 1 2 LOGARITMOS csch x = = ∑ k 2 = 6 ( 2n3 + 3n2 + n ) n 1 Gráfica 2. Las funciones trigonométricas csc x , log a N = x ⇒ a x = N tg (θ + nπ ) = tg θ senh x e x − e − x k =1 sec x , ctg x : log a MN = log a M + log a N senh : → sen ( nπ ) = 0 ∑ k 3 = 4 ( n 4 + 2n3 + n 2 ) n 1 M 2.5 cosh : → [1, ∞ = log a M − log a N cos ( nπ ) = ( −1) n log a k =1 2 N tgh : → −1,1 ∑ k 4 = 30 ( 6n5 + 15n4 + 10n3 − n ) n 1 tg ( nπ ) = 0 1.5 log a N r = r log a N 1 ctgh : − {0} → −∞ , −1 ∪ 1, ∞ k =1 ⎛ 2n + 1 ⎞ + ( 2n − 1) = n 2 π ⎟ = ( −1) → 0,1] 0.5 n log b N ln N 1+ 3 + 5 + sen ⎜ log a N = = sech : ⎝ 2 ⎠ 0 log b a ln a n − {0} → − {0} n! = ∏ k -0.5 csch : log10 N = log N y log e N = ln N ⎛ 2n + 1 ⎞ k =1 -1 cos ⎜ π⎟=0 ALGUNOS PRODUCTOS -1.5 ⎝ 2 ⎠ Gráfica 5. Las funciones hiperbólicas senh x , ⎛n⎞ n! csc x a ⋅ ( c + d ) = ac + ad ⎜ ⎟= , k≤n -2 sec x ⎛ 2n + 1 ⎞ cosh x , tgh x : ⎝ k ⎠ ( n − k )!k ! ctg x tg ⎜ π⎟=∞ ⎝ 2 ⎠ -2.5 ( a + b) ⋅ ( a − b) = a − b -8 -6 -4 -2 0 2 4 6 8 2 2 5 n ⎛n⎞ ( x + y ) = ∑ ⎜ ⎟ xn−k y k π⎞ n Gráfica 3. Las funciones trigonométricas inversas ⎛ ( a + b ) ⋅ ( a + b ) = ( a + b ) = a 2 + 2ab + b2 4 sen θ = cos ⎜θ − ⎟ 2 k =0 ⎝ k ⎠ arcsen x , arccos x , arctg x : ⎝ 2⎠ 3 ( a − b ) ⋅ ( a − b ) = ( a − b ) = a 2 − 2ab + b 2 2 2 ( x1 + x2 + + xk ) n =∑ n! x1n1 ⋅ x2 2 ⎛ π⎞ cosθ = sen ⎜θ + ⎟ n nk x 4 ( x + b ) ⋅ ( x + d ) = x 2 + ( b + d ) x + bd 1 k n1 !n2 ! nk ! 3 ⎝ 2⎠ 0 ( ax + b ) ⋅ ( cx + d ) = acx 2 + ( ad + bc ) x + bd CONSTANTES sen (α ± β ) = sen α cos β ± cos α sen β -1 ( a + b ) ⋅ ( c + d ) = ac + ad + bc + bd π = 3.14159265359… 2 cos (α ± β ) = cos α cos β ∓ sen α sen β -2 se nh x e = 2.71828182846… -3 co sh x ( a + b ) = a3 + 3a 2b + 3ab2 + b3 tg α ± tg β 3 1 tgh x tg (α ± β ) = -4 -5 0 5 TRIGONOMETRÍA 1 ∓ tg α tg β ( a − b ) = a3 − 3a 2b + 3ab2 − b3 3 0 CO 1 FUNCIONES HIPERBÓLICAS INV sen θ = cscθ = sen 2θ = 2sen θ cosθ ( a + b + c ) = a 2 + b2 + c 2 + 2ab + 2ac + 2bc 2 HIP CA sen θ 1 -1 arc sen x arc cos x arc tg x cos 2θ = cos 2 θ − sen 2 θ ( senh −1 x = ln x + x 2 + 1 , ∀x ∈ ) cosθ = secθ = ( a − b ) ⋅ ( a + ab + b ) = a − b ( ) -2 2 tg θ 2 2 3 3 cosθ -3 -2 -1 0 1 2 3 HIP tg 2θ = cosh −1 x = ln x ± x 2 − 1 , x ≥ 1 sen θ CO 1 − tg 2 θ ( a − b ) ⋅ ( a3 + a 2 b + ab2 + b3 ) = a 4 − b4 tg θ = = ctg θ = 1 1 ⎛1+ x ⎞ cosθ CA tg θ 1 tgh −1 x = ln ⎜ ⎟, x < 1 ( a − b ) ⋅ ( a 4 + a3b + a 2 b2 + ab3 + b 4 ) = a5 − b5 sen 2 θ = (1 − cos 2θ ) 2 ⎝ 1− x ⎠ 2 ⎛ ⎞ π radianes=180 1 ⎛ x +1 ⎞ ctgh −1 x = ln ⎜ n 1 ( a − b ) ⋅ ⎜ ∑ a n − k b k −1 ⎟ = a n − b n ∀n ∈ cos 2 θ = (1 + cos 2θ ) ⎟, x > 1 2 ⎝ x −1 ⎠ ⎝ k =1 ⎠ 2 1 − cos 2θ ⎛ 1 ± 1 − x2 ⎞ HIP tg 2 θ = sech −1 x = ln ⎜ ⎟, 0 < x ≤ 1 CO 1 + cos 2θ ⎜ x ⎟ ⎝ ⎠ θ ⎛1 x2 + 1 ⎞ csch −1 x = ln ⎜ + ⎟, x ≠ 0 CA ⎜x x ⎟ ⎝ ⎠