Hsn course revision notes

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Hsn course revision notes

  1. 1. hsn.uk.net Higher Mathematics HSN24400 Course Revi on Notes si Thi docum ent was produced speci f the HSN.uk.net websi and we requi that any copi or deri ve s ally or te, re es vati works attribute the work to us. For m ore detai about the copyri on these notes, please see http://creati ls ght vecom m ons.org/licenses/by-nc-sa/2.0/
  2. 2. Higher Mathematics Course Revision Notes Contents Unit 1 – Mathematics 1 Straight Lines 1 Functions and Graphs 2 Differentiation 5 Sequences 6 Unit 2 – Mathematics 2 Polynomials and Quadratics 7 Integration 8 Trigonometry 9 Circles 12 Unit 3 – Mathematics 3 Vectors 12 Further Calculus 15 Exponentials and Logarithms 16 The Wave Function 18 hsn.uk.net - ii - HSN24400
  3. 3. Higher Mathematics Course Revision Notes Straight Lines Distance Formula stance = ( x2 − x1 ) + ( y2 − y1 ) between poi ( x1, y1 ) and ( x2 , y2 ) 2 2 Di nts Gradients y2 − y1 m= between the poi ( x1, y1 ) and ( x2 , y2 ) where x1 ≠ x2 nts x 2 − x1 Posi ve gradi ti ents, negati gradi ve ents, zero gradients, undef ned gradi i ents eg x = 2 eg y = 4 Li wi the sam e gradi are parallel nes th ent eg The li parallel to 2 y + 3 x = 5 ne has gradi m = − 3 si ent 2 nce 2 y + 3 x = 5 2 y = −3 x + 5 5 y = − 3 x + 2 (m ust be i the f 2 n orm y = mx + c ) Perpendicular li have gradi nes ents such that m × mperp. = −1 eg i m = 2 then mperp. = − 2 f 3 3 m = tan y i the angle that the li m akes wi s ne th positive direction the posi ve di on of the x-axi ti recti s x Equation of a Straight Line The li passi through ( a, b ) wi gradi m has equati ne ng th ent on: y − b = m( x − a ) Medians B x1 + x 2 y1 + y2 M i the m i s dpoi of AC, i M = nt e , 2 2 BM i not usually perpendi s cular to AC, so m1 × m2 = −1 A C cannot be used M To work out the gradi of BM, use the gradi f ula ent ent orm hsn.uk.net Page 1 HSN24400
  4. 4. Higher Mathematics Course Revision Notes Altitudes B D i not usually the m i s dpoi of AC nt BD i perpendi s cular t AC, so m1 × m2 = −1 can be used t o o A C work out the gradi of BD ent D Perpendicular Bisectors D CD passes through m i dpoi of AC nt A B C CD i perpendi s cular t AB, so m1 × m2 = −1 can be used t o o C B f nd the gradi of CD i ent Perpendicular bisectors do not necessari have to appear ly D wi n a tri thi angle – they can occur wi st ght li th rai nes A Functions and Graphs Composite Functions Example If f ( x ) = x 2 − 2 and g ( x ) = 1 , f nd a f ula f x i orm or (a) h ( x ) = f ( g ( x ) ) (b) k ( x ) = g ( f ( x ) ) and state a suitable dom ai f each. n or (a) h ( x ) = f ( g ( x ) ) (b) k ( x ) = g ( f ( x ) ) = f (1) x = g( x 2 − 2) =(1) −2 2 1 x = x −2 2 1 = −2 Dom ai { x : x ∈ , x ≠ ± 2 } n: x2 Dom ai { x : x ∈ , x ≠ 0} n: You wi probably only be asked f a dom ai i the f ll or n f uncti i on nvolved a f on or an racti even root. Rem em ber that i a f on the denom i n racti nator cannot be zero and any num ber bei square rooted cannot be negati ng ve eg f ( x ) = x + 1 could have dom ai { x : x ∈ , x ≥ −1} n: hsn.uk.net Page 2 HSN24400
  5. 5. Higher Mathematics Course Revision Notes Graphs of Inverses To draw the graph of an inverse functi ref on, lect the graph of the functi i the on n li y = x ne y y=x y = g( x ) y = g−1 ( x ) O x Exponential and Logarithmic Functions Exponential Logarithmic y y y y = loga x y = ax, a >1 y = ax, 0 < a <1 ( a ,1) O x 1 (1, a ) 1 (1, a ) 1 O x O x Trigonometric Functions y = si x n y = cos x y = tan x y y y 1 1 O x O x O x –1 –1 180° 360° 180° 360° 180° 360° Peri = 360° od Peri = 360° od Peri = 180° od tude = 1 Am pli tude = 1 Am pli Am pli tude i undef ned s i Graph Transformations The next page shows the ef ect of transf ati on the two graphs shown below. f orm ons ( 2, 2 ) y = g ( x ) y y 1 O x –1 0 3 x –1 180° 360° hsn.uk.net Page 3 HSN24400
  6. 6. Higher Mathematics Course Revision Notes Function Ef ect f Ef ect on f ( x ) f Ef ect on si x ° f n f ( x ) + a Shi ts the graph f y y = g( x ) + 1 y y = si x ° + 1 n a up the ( 2, 3 ) • 2 y-axis 1 • • ( 3,1) ( −1,1) 0 x 0 x 180° 360° f ( x + a ) Shi ts the graph f y y = g ( x + 1) y y = si ( x − 90 ) ° n −a along the • (1, 2 ) 1 x-axis 0 x –2 0 2 x –1 180° 360° − f (x) Reflects the y y y = − si x ° n y = − g( x ) graph i the n 0 1 x-axis –1 3 x 0 x • ( 2, − 2 ) –1 180° 360° f (−x ) Ref lects the ( −2, 2 ) y y = g( −x ) y = si ( − x ° ) n y graph i the n • 1 y-axi s 0 x –3 0 1 x –1 −360° −180° kf ( x ) Scales the graph y ( 2, 4 ) y = 2 g ( x ) y y = 1 si x ° 2 n vertically • 1 Stretches if 2 k >1 −1 0 x x 2 Com presses if –1 0 3 k <1 180° 360° f ( kx ) Scales the graph y (1, 2 ) y = g(2x ) y y = si 2 x ° n horizontally • 1 Com presses if k >1 0 x Stretches if x −1 0 3 –1 k <1 2 2 180° 360° hsn.uk.net Page 4 HSN24400
  7. 7. Higher Mathematics Course Revision Notes Differentiation Differentiating If f ( x ) = ax n then f ′ ( x ) = anx n −1 Bef you di f ore f erenti all brackets should be m ulti ed out, and there should be no ate, pli f ons wi an x term i the denom i racti th n nator (bottom li f exam ple: ne), or 3 1 −1 1 = 1 x −2 = 3x −2 =x 2 3x 2 3 x2 x Equations of Tangents Tangents are strai li ght nes, theref to f nd the equati of a tangent, you need a poi ore i on nt on the li and i gradi to substi ne ts ent tute i y − b = m ( x − a ) nto You wi always be gi one coordi ll ven nate of the poi whi the tangent touches nt ch Fi the other coordi nd nate by solvi the equati of the curve ng on nd ent f erenti ng then substi ng i the x-coordi Fi the gradi by di f ati tuti n nate of the point Example Fi the equati of the tangent to the graph of y = x 3 at the poi where x = 9 . nd on nt y = x3 y = x3 At x = 9, m = 3 × 9 2 1 y − b = m(x − a) 2 = 93 = x2 3 =3 9 y − 27 = 9 ( x − 9 ) 2 2 = 3 ×3 2 y − 54 = 9 x − 81 = 33 dy 3 1 = 2 x2 2 2 y = 9 x − 27 = 27 dx =9 2 ( 9, 27 ) dy Stationary poi occur at poi where nts nts =0 dx You m ust j f the nature of turni poi or poi of i lecti usti y ng nts nts nf on hsn.uk.net Page 5 HSN24400
  8. 8. Higher Mathematics Course Revision Notes Graphs of Derived Functions Quadratic Cubic Quartic y y y 0 0 0 – + + – + – + – 0 + x x x 0 0 y y y + + + x – – x – + + x – Linear Quadratic Cubic Optimisation These types of questions are usually practi problem s whi i cal ch nvolve m axi um or m m i m um areas or volum es ni Rem em ber you m ust show that a m axi um or m i m um exi m ni sts Sequences Linear Recurrence Relations A linear recurrence relati i i the f on s n orm un+1 = aun + b . Also be aware that thi m ay be s written as un = aun−1 + b b If −1 < a < 1 then a li i l = m t exi You m ust state thi whenever you use the li i sts. s m t 1− a f ula orm hsn.uk.net Page 6 HSN24400
  9. 9. Higher Mathematics Course Revision Notes Polynomials and Quadratics Polynomials The degree of a polynom i i the value of the hi al s ghest power, eg 3 x 4 + 3 has degree 4 Syntheti di si (nested f ) can be used to f c vi on orm actori polynom i se als Example 4 x 3 − 7 x 2 + 11 Find . x +2 –2 4 –7 0 11 Rem em ber to put i 0 n i there i no term f s –8 30 –60 4 –15 30 –49 4 x 3 − 7 x 2 + 11 = 4 x 2 − 15 x + 30 rem ainder − 49 x +2 i 4 x 3 − 7 x 2 + 11 = ( x + 2 ) ( 4 x 2 − 15 x + 30 ) − 49 e If the di sor i a f vi s actor then the rem ainder i zero s If the rem ainder i zero then the di sor i a f s vi s actor Completing the Square The x 2 term m ust have a coef i ent of one. If i does not, you m ust take out a com m on f ci t 2 factor from the x and x term , but not the constant orm y = a ( x + p ) + q the turni poi of the graph i ( − p, q ) 2 In the f ng nt s Example Wri 3 x 2 − 12 x + 7 i the f te n orm a(x + p)2 + q . 3 x 2 − 12 x + 7 = 3( x 2 − 4x ) + 7 = 3 ( x 2 − 4 x + ( −2 )2 − ( −2 )2 ) + 7 = 3 ( ( x − 2 )2 − 4 ) + 7 = 3 ( x − 2 )2 − 12 + 7 = 3 ( x − 2 )2 − 5 Note that i thi exam ple, the graph i ∪ -shaped si the x 2 coef i ent i posi ve; and n s s nce f ci s ti the turni poi i ( 2, − 5 ) . ng nt s hsn.uk.net Page 7 HSN24400
  10. 10. Higher Mathematics Course Revision Notes The Discriminant The di m i scri nant i part of the quadrati f ula and can be used to i cate how m any s c orm ndi roots a quadrati has. For the quadrati ax + bx + c : c c 2 If b 2 − 4 ac > 0 , the roots are real and unequal (di nct) sti 2 roots If b 2 − 4 ac = 0 , the roots are real and equal (i repeated roots) e 1 root If b 2 − 4 ac < 0 , the roots are not real; they do not exist no roots The di m i scri nant can also be used to calculate the num ber of intersecti between a ons li and a curve. To use i you m ust f rst equate them and set equal to zero, bef ne t, i ore usi the di m i ng scri nant Rem em ber i b 2 − 4 ac = 0 , the li i a tangent f ne s Integration Integrating ax n +1 ax n dx = +c n +1 As wi di f th f erenti on, all brackets m ust be m ulti ed out, and there m ust be no ati pli f ons wi an x term i the denom i racti th n nator Examples dx x 2 + 5x 7 1. Find . 2. dx 8 x5 x2 dx 1 x 2 + 5x 7 = dx 2 dx = x −2 ( x 2 + 5 x 7 ) dx 8 x5 8 x5 x = x −5 8 dx = x 0 + 5 x 5 dx x8 3 = 1 + 5 x 5 dx = 3 +c 8 = x + 5 x6 + c 6 3 = 8 x8 + c 3 = 3 8 x3 + c 8 hsn.uk.net Page 8 HSN24400
  11. 11. Higher Mathematics Course Revision Notes The Area under a Curve b If F ( x ) i the i s ntegral of f ( x ) , then f ( x ) dx = F (b ) − F ( a ) a y y = f (x) a b x Rem em ber that areas spli by the x-axi m ust be calculated separately and any negati t s ve si i gns gnored; these j show that the area i under the axi ust s s. The Area between two Curves b The area between the graphs of y = f ( x ) and y = g ( x ) i def ned as s i a f ( x ) − g ( x ) dx y y = g( x ) ven, f ( x ) and g ( x ) If the li i are not gi m ts should be equated to f nd a and b i x a b y = f (x) Trigonometry Background Knowledge You should know how to use all of the i orm ati below: nf on SOH CAH TOA si x n tan x = cos x si 2 x + cos2 x = 1 n a b c The si rule: ne = = si A si B si C n n n b2 + c 2 − a2 The cosi rule: a = b + c − 2bc cos A or cos A = ne 2 2 2 2bc hsn.uk.net Page 9 HSN24400
  12. 12. Higher Mathematics Course Revision Notes The area of a triangle, A = 1 ab si C 2 n CAST diagram s Exact values: 2 30° 2 45° 3 1 45° 60° 1 1 Radians You should know how to convert between radi and degrees: ans 360° = 2 90° = 2 45° = 4 ÷ 180 × → Degrees  Radians 180° = 60° = 3 30° = 6 ans × 180 ÷ → Radi  Degrees 5 × 180 eg 5 = 6 = 150° 6 Trigonometric Equations Look at the restri ons on the dom ai eg 0 ≤ x ° < 360, or 0 ≤ x < cti n, Be aware of whether the answer i requi i degrees or radi s red n ans Rem em ber a CAST diagram whenever you are asked to “solve” Examples 1. Solve 3si 2 x° = 1 where 0 ≤ x ° < 360 . n 3si 2 x ° = 1 n 3 ( si x ° ) = 1 2 n ( si x ° )2 = 1 n 3 si x ° = ± 1 n 3 S A x ° = si n −1 (± 1 ) 3 T C x° = 35.3° x° = 180 − 35.3 x° = 180 + 35.3 x° = 360 − 35.3 = 144.7° = 215.3° = 324.7° Soluti set = {35.3°, 144.7°, 215.3°, 324.7°} on hsn.uk.net Page 10 HSN24400
  13. 13. Higher Mathematics Course Revision Notes 2. Solve 2 si 2 x − 1 = 0 , 0 ≤ x < 2 . n 2 si 2 x − 1 = 0 n 2si 2 x = 1 n si 2 x = 1 n 2 S A 2 x = si −1 ( 1 ) n 2 T C 2 x ° = 30° 2 x ° = 180° − 30° x ° = 15° 2 x ° = 150° x ° = 75° 2 x ° = 360° + 30° 2 x ° = 360° +180° − 30° 2 x ° = 390° 2 x ° = 510° x ° = 195° x ° = 255° 15 15° = 180 75 75° = 180 195 195° = 180 255 255° = 180 = 36 3 = 15 36 = 36 39 51 = 36 = 12 5 = 12 = 12 13 = 17 12 ons { Soluti set = 12 , 12 , 12 , 17 5 13 12 } Compound Angle Formulae cos ( A ± B ) = cos A cos B si A si B n n si ( A ± B ) = si A cos B ± cos A si B n n n These are gi on the f ula sheet ven orm Double Angle Formulae si 2 A = 2si A cos A n n cos 2 A = cos2 A − si 2 A n = 1 − 2 si 2 A n = 2 cos 2 A − 1 These are gi on the f ula sheet ven orm hsn.uk.net Page 11 HSN24400
  14. 14. Higher Mathematics Course Revision Notes Circles Equations of Circles A ci wi centre ( a,b ) and radi r has the equati ( x − a )2 + ( y − b ) = r2 2 rcle th us on Note that i a ci has centre ( 0, 0 ) then the equati i x 2 + y 2 = r2 f rcle on s The equati can also be gi i the f on ven n orm x 2 + y 2 + 2 gx + 2 fy + c = 0 where the centre i ( − g,− f s ) and the radi r = g2 + f 2 − c us You do not have to rem em ber any of these equations, si they are all gi i the nce ven n exam stance f ula, d = ( x2 − x1 ) + ( y2 − y1 ) , si thi 2 2 You wi have to rem em ber the di ll orm nce s i not gi s ven, and i f s requently used i ci questi n rcle ons Intersection of a Line and a Circle two intersections one intersection (tangency) no intersections Rem em ber, a tangent and a li f ne rom the centre of a circle wi m eet at ri angles, ll ght whi m eans that m1 × m2 = −1 can be used ch Vectors Basic Facts A vector i a quanti wi both m agni s ty th tude (si and di on ze) recti A vector i nam ed ei s ther by usi a di ng rected li segm ent (eg AB ) or a bold letter ne (eg u written u) A vector m ay also be def ned i term s of i, j and k, the uni vectors i three i n t n perpendicular di ons: recti 1 0 0 i= 0 j= 1 k= 0 0 0 1 a The m agnitude of vector AB = b i def ned as AB = a 2 + b 2 s i hsn.uk.net Page 12 HSN24400
  15. 15. Higher Mathematics Course Revision Notes a1 b1 a1 ± b1 a1 ka1 0 a2 ± b2 = a2 ± b2 k a2 = ka2 where k i a scalar s Z ero vector: 0 a3 b3 a3 ± b3 a3 ka3 0 OA i called the posi on vector of the poi A relati to the ori n, wri s ti nt ve gi tten a AB = b − a where a and b are the posi on vectors of A and B ti If AB = k BC where k i a scalar, then AB i parallel to BC . Si B i com m on to both s s nce s AB and k BC , then A, B and C are collinear Dividing Vectors in a Ratio The poi P can also be worked out f nt rom f rst pri ples, or i nci U si the secti f ula. If P di des AB i the rati m : n , then: ng on orm vi n o n m p= a+ b where p i the posi on vector OP s ti m+n m+n Example P i the poi ( −2, 4, − 1) and R i the poi ( 8, − 1,19 ) . Poi T di des PR i the rati s nt s nt nt vi n o 2:3. Work out the coordinates of poi T. nt 2:3 P R T U si the secti f ul ng on orm a From fi pri pl rst nci es The rati i 2:3, so let m = 2 and n = 3 o s PT 2 = n m TR 3 t= p+ r m+n m+n 3PT = 2TR = 3 p + 2r 5 5 ( ) 3 t − p = 2 (r − t ) ( = 1 3 p + 2r 5 ) 3t − 3 p = 2r − 2t 3t + 2t = 2r + 3 p −6 16 =1 12 + −2 16 −6 5 −3 38 5t = −2 + 12 38 −3 10 = 1 10 10 5 35 5t = 10 35 2 = 2 2 7 t= 2 7 Theref T i the poi ( 2, 2, 7 ) . ore s nt Theref T i the poi ( 2, 2, 7 ) . ore s nt hsn.uk.net Page 13 HSN24400
  16. 16. Higher Mathematics Course Revision Notes The Scalar Product The scalar product a.b = a b cos , where i the sm allest angle between a and b s Rem em ber that both vectors m ust poi away f nt rom the angle, eg a b a1 b1 If a = a2 and b = b2 then a.b = a1b1 + a2b2 + a3b3 a3 b3 a.b ab +a b +a b cos = or cos = 1 1 2 2 3 3 ab a b If a and b are perpendicular then a.b = 0 If a.b = 0 then a and b are perpendicular Example 8 4 If u = 0 and v = 0 , calculate the angle between the vectors u + v and u − v . 4 1 Let a = u + v Let b = u − v 8 4 8 4 a= 0 + 0 b= 0 − 0 4 1 4 1 12 4 a= 0 b= 0 5 3 a.b cos = ab (12 × 4 ) + ( 0 × 0 ) + ( 5 × 3 ) = 122 + 02 + 52 4 2 + 02 + 32 63 = 169 25 63 = cos −1 169 25 = 14· ° 3 hsn.uk.net Page 14 HSN24400
  17. 17. Higher Mathematics Course Revision Notes Further Calculus Trigonometry Differenti on ati Thi i strai orward, si the f ulae are gi on the f ula sheet: s s ghtf nce orm ven orm f (x) f ′( x ) si ax n a cos ax cos ax − a si ax n Integration Agai the f ulae are provi i the paper: n, orm ded n f (x) f ( x ) dx si ax n − 1 cos ax + c a cos ax 1 si ax + c a n Examples f erenti x 3 + cos3 x wi respect to x. 1. Di f ate th ( ) d x 3 + cos3 x = 3 x 2 − 3si x dx n3 2. Find 4 x 3 + si x dx . n3 4x 4 1 4 x + si x dx = 3 n3 − 3 cos3 x + c 4 = x 4 − 1 cos3 x + c 3 Chain Rule Differentiation n −1 n −1 If f ( x ) = ( ax + b )n then f ′ ( x ) = n ( ax + b ) × a = an ( ax + b ) or n −1 If f ( x ) = ( p ( x ) ) then f ′ ( x ) = n ( p ( x ) ) n × p′ ( x ) “The power m ulti es to the f pli ront, the bracket stays the sam e, the power lowers by one and everythi i m ulti ed by the di f ng s pli f erenti of the bracket” al hsn.uk.net Page 15 HSN24400
  18. 18. Higher Mathematics Course Revision Notes Examples 1 1. G i f ( x ) = ven + x − si x , f nd f ′ ( x ) . n3 i x2 1 f ( x ) = x −2 + x 2 − si x n3 −1 f ′ ( x ) = −2 x −3 + 1 x 2 − 3cos3 x 2 2 1 =− 3 + − 3cos3 x x 2 x ven f ( x ) = ( 3 x 2 + 2 x + 1) , f nd f ′ ( x ) . 3 2. G i i f ′ ( x ) = 3 ( 3 x 2 + 2 x + 1) × ( 6 x + 2 ) 2 = 3 ( 6 x + 2 ) ( 3 x 2 + 2 x + 1) 2 f erenti y = cos 2 x = ( cos x )2 wi respect to x. 3. Di f ate th dy = 2 ( cos x ) × ( − si x ) n dx = −2cos x si x n n Integration of (ax + b) ( ax + b )n+1 ( ax + b ) dx = n +c ( n + 1) × a Example Find ( 3 x + 5 )4 dx . ( 3 x + 5 )5 ( 3 x + 5 )5 ( 3 x + 5 ) dx = 4 +c = +c 5× 3 15 It i possi f any type of ‘urther calculus’ to be exam i i the style of a standard s ble or f ned n calculus questi (eg opti i on, area under a curve, etc) on m sati Exponentials and Logarithms An exponenti i a f al s uncti i the f on n orm f ( x ) = a x Logari s and exponenti are i thm als nverses y = a x ⇔ loga y = x On a calculator, log i log10 and ln i loge s s hsn.uk.net Page 16 HSN24400
  19. 19. Higher Mathematics Course Revision Notes Laws of Logarithms loga x + loga y = loga xy (Squash) loga x − loga y = loga x y (Split) loga x n = n loga x (Fly) Examples 1. Evaluate log2 4 + log2 6 − log2 3 log2 4 + log2 6 − log2 3 4×6 = log2 3 = log2 8 =3 (si 23 = 8) nce 2. Below i a di s agram of part of the graph of y = ke 0.7 x y y = ke 0.7 x 3 0 x =1 x (a) Fi the value of k nd (b) The li wi equati x = 1 i ne th on ntersects at R. Fi the coordi nd nates of R. (a) At ( 0, 3 ) , y = ke 0.7 x 3 = ke 0.7×0 3 = ke 0 k =3 (b) x = 1 y = 3e 0.7×1 = 6.04 So R i the poi (1, 6.04 ) . s nt hsn.uk.net Page 17 HSN24400
  20. 20. Higher Mathematics Course Revision Notes The Wave Function Example 1. Express 6 si x ° − 2 cos x ° i the f n n orm k cos ( x − a ) ° where 0 ≤ a ° < 360 . k cos ( x − a ) ° = k cos x ° cos a ° + k si x ° si a ° n n = k cos a ° cos x ° + k si a ° si x ° n n k si a ° k cos a ° = − 2 (− 2 ) n 2 2 k= + 6 tan a ° = k cos a ° k si a ° = 6 n = 2+6 6 =− S A = 8 2 T C =2 2 =− 3 a ° = 180° − tan −1 ( 3 ) = 180° − 60° = 120° Therefore 6 si x ° − 2 cos x ° = 2 2 cos ( x − 120 ) ° . n 2. Express cos x − si x i the f n n orm k si ( x + ) where 0 ≤ < 2 . n k si ( x + ) = k si x cos + k cos x si n n n = k cos si x + k si cos x n n k cos = −1 k = ( −1)2 + 12 k si n tan = k si = 1 n = 2 k cos = −1 S A ° = 180° − tan −1 (1) T C = 180° − 45° = 135° 135 = 180 =3 4 Theref cos x − si x = 2 si x + 3 ore n n 4 ( ). The m axi um value of an expressi i the f m on n orm k cos ( x ± a ) occurs when cos ( x ± a ) = 1 ; and si ( x ± a ) = 1 f k si ( x ± a ) n or n The m i m um value of an expressi ni on i the f n orm k cos ( x ± a ) occurs when cos ( x ± a ) = −1 ; and si ( x ± a ) = −1 f k si ( x ± a ) n or n hsn.uk.net Page 18 HSN24400

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