additionalmathematicstrigonometricf
unctionsadditionalmathematicstrigo
nometricfunctionsadditionalmathem
aticstrigonometri...
zefry@sas.edu.my 2
TRIGONOMETRIC FUNCTIONS
5.1 Positive Angle and Negative Angle
Positive Angle Negative Angle
Represent e...
zefry@sas.edu.my 3
5.2 Six Trigonometric Functions of any Angle (1)
5.2.1 Define sine, cosine and tangent of any angle in ...
zefry@sas.edu.my 4
(c)  in quadrant III
OP =
=
sin  =
 
3
cos  =
 
4
tan  =
 
3
=
(d)  in quadrant IV
OP =
=...
zefry@sas.edu.my 5
(e) Conclusion:
Reference angle (RA) is the acute angle formed between the rotating ray of the angle an...
zefry@sas.edu.my 6
5.2.2 Define cotangent, secant and cosecant of any angle in a Cartesian plane.
1
 
 
 
 
 
 ...
zefry@sas.edu.my 7
5.2.3 Find values of six trigonometric functions of any angle
1. Complete the table below.
30 45 60
...
zefry@sas.edu.my 8
5.2.4 Solve trigonometric equations
A. Steps to solve trigonometric equation
1. Determine the range of ...
zefry@sas.edu.my 9
d. sin  =  0.9397 e. tan  =  0.3640
Range :
Reference angle :
Quadrant :
Quadrant ___ Quadrant ___ ...
zefry@sas.edu.my 10
b. cos 3 =  0.9781
c. tan
2

=  2.05
Range
Reference angle :
Quadrant :
Actual angles
d. sin ( + ...
zefry@sas.edu.my 11
j. sin  = cos 20 k. cos  =  sin 55
Example : 2 sin x cos x = cos x
2 sin x cos x  cos x = 0
cos ...
zefry@sas.edu.my 12
3. Given that px sin and 00
< x < 900
.
Express each of the following trigonometric ratios in terms o...
zefry@sas.edu.my 13
4. Given that
17
8
sin x and 2700
< x < 3600
.
Without using tables or calculator, find the values
o...
zefry@sas.edu.my 14
5.4 Basic Identities
5.4.1 Prove Trigonometric Identities using Basic Identities
Three basic trigonome...
zefry@sas.edu.my 15
d.  2 1
1
sin
sec tan
sin
 
   
 
e.
1
2
1
sin x cos x
sec x
cos x sin x

 

2. Solve th...
zefry@sas.edu.my 16
ANSWERS
5.2.1
5. cos 51= 0.6293
5.2.4
2a. 0  2  720 , 54.33
 = 27.17, 62.83, 207.17, 242....
zefry@sas.edu.my 17
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Trigo functions

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Trigo functions

  1. 1. additionalmathematicstrigonometricf unctionsadditionalmathematicstrigo nometricfunctionsadditionalmathem aticstrigonometricfunctionsadditiona lmathematicstrigonometricfunctions additionalmathematicstrigonometricf unctionsadditionalmathematicstrigo nometricfunctionsadditionalmathem aticstrigonometricfunctionsadditiona lmathematicstrigonometricfunctions additionalmathematicstrigonometricf unctionsadditionalmathematicstrigo nometricfunctionsadditionalmathem aticstrigonometricfunctionsadditiona lmathematicstrigonometricfunctions additionalmathematicstrigonometricf unctionsadditionalmathematicstrigo nometricfunctionsadditionalmathem aticstrigonometricfunctionsadditiona lmathematicstrigonometricfunctions TRIGONOMETRIC FUNCTIONS Name ........................................................................................
  2. 2. zefry@sas.edu.my 2 TRIGONOMETRIC FUNCTIONS 5.1 Positive Angle and Negative Angle Positive Angle Negative Angle Represent each of the following angles in a Cartesian plane and state the quadrant of the angle. Example 60 Quadrant 1 1(a) 70 (b) 150 Example 215 Quadrant III 2(a) 195 (b) 345 Example 395 Quadrant I 3(a) 415 (b) 480 Example 5 4  Quadrant III 4(a) 3 4  (b) 5 3  Example 45 Quadrant IV 5(a) 130 (b) 1 3   2 radian = 360  radian = 180 60 y xO y xO y xO Quadrant II Quadrant I Quadrant III Quadrant IV y x 90 2       0 360 (2) 180 () 270 3 2       215 y xO y xO y xO A positive angle is measured in an anticlockwise direction from the positive x-axis. 45 y xO y xO y xO y xO y xO 5 4  y xO y xO y xO 60 y xO Anticlockwise direction 45 y xO Clockwise direction 395 y xO 35 360 A negative angle is measured in a clockwise direction from the positive x-axis.
  3. 3. zefry@sas.edu.my 3 5.2 Six Trigonometric Functions of any Angle (1) 5.2.1 Define sine, cosine and tangent of any angle in a Cartesian plane 1             sin cos tan     Conclusion : 2 r2 = 32 + 42 r = 2 2 3 4 r = 5 Conclusion : Pythagoras’ Theorem : 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 c a b c a b a c b a c b b c a b c a                3. Find the length of OA and the values of sine, cosine and tangent of . (a)  in quadrant I OP = = sin  =   5 cos  =   12 tan  =   5 (b)  in quadrant II OP = = sin  =   6 = cos  =   8 = tan  =   6 =  x r y Opposite sin Hypotenuse Adjacent cos Hypotenuse Opposite tan Adjacent        3 r 4 b ca P (12, 5) 5 12O x y  P (8, 6) x y O  8 6 Opposite to  Hypotenuse Adjacent to  
  4. 4. zefry@sas.edu.my 4 (c)  in quadrant III OP = = sin  =   3 cos  =   4 tan  =   3 = (d)  in quadrant IV OP = = sin  =   12 cos  =   5 tan  =   12 (e) Conclusion: Sin  is positive for  in quadrant ……. and ……. Cos  is positive for  in quadrant ……. and ……. Tan  is positive for  in quadrant ……. and ……. Sin  is negative for  in quadrant ……. and ……. Cos  is negative for  in quadrant ……. and ……. Tan  is negative for  in quadrant ……. and ……. 4. Find the corresponding reference angle of . (a) Reference angle = 55 (b) Reference angle =  110 = 70 (c) Reference angle = 215  = 35 (d) Reference angle =  300 = 60 y x 90 0 360 180 270 Sin  Cos  Tan  Sin  Cos  Tan  Sin  Cos  Tan  + Sin  Cos  Tan  P (4, 3) x y 4 O  3 P (5, 12) x y 5 O  12 55 y x Fill in with  or + sign. 110 y x180 360 215 y x180 360 300 y x180 360
  5. 5. zefry@sas.edu.my 5 (e) Conclusion: Reference angle (RA) is the acute angle formed between the rotating ray of the angle and the ______________________________ In Quadrant II: In Quadrant III In Quadrant IV sin  = sin (180  ) sin  = sin (  180) sin  = sin (360  ) cos  = cos (180  ) cos  = cos (  180) cos  = cos (360  ) tan  = tan (180  ) tan  = tan (  180) tan  = tan (360  ) 5. Given that cos 51 = 0.6293, find the trigonometric ratios of cos 231 without using a calculator or mathematical tables. Reference angle of 231 = 231  = cos 231 = = 6. Given that sin 70 = 0.9397, find the trigonometric ratios of sin 610 without using a calculator or mathematical tables. Reference angle of 610 = 610   = sin 610 = = 7. Given that tan 25 = 0.4663, find the trigonometric ratios of tan 335 without using a calculator or mathematical tables. Reference angle of 335 =  335 = tan 335 = = R.A = R.A =   R.A =   R.A =   y x y xO 180  y xO 180    y x O 360    y x y x y x
  6. 6. zefry@sas.edu.my 6 5.2.2 Define cotangent, secant and cosecant of any angle in a Cartesian plane. 1             sin cos tan       2                     1 1 r sin y 1 1 cos 1 1 tan          3. Definition of cotangent , secant  and cosecant . 1 cosec sin 1 sec cos 1 cot tan          4. Since sin tan cos     , then cot   5.       sin sin 90 cos cos 90 tan tan 90 y x r r x y r r y x x y                6. Complementary angles: sin  = cos (90  ) cos  = sin (90  ) tan  = cot (90  ) cosec  = sec (90  ) sec  = cosec (90  ) cot  = tan (90  ) 7. Given that sin 48 = 0.7431, cos 48 = 0.6991 and tan 48 = 1.1106, evaluate the value of cos 42. cos 42 = = = 8. Given that sin 67 = 0.9205, cos 67 = 0.3907 and tan 67 = 2.3559, evaluate the value of cot 23. cot 23 = = = 9. Given that sin 37 = 0.6018, cos 37 = 0.7986 and tan 37 = 0.7536, evaluate the value of sec 53. sec 53 = = =  x r y  x r y 90   x r y 48 90  48 67 90  67 37 90  37
  7. 7. zefry@sas.edu.my 7 5.2.3 Find values of six trigonometric functions of any angle 1. Complete the table below. 30 45 60 sin  1 2 cos  1 2 tan  1 2. Use the values of trigonometric ratio for the special angles, 30, 45 and 60, to find the value of the trigonometric functions below Example: Evaluate sin 210 a. Evaluate tan 300 Draw diagram to determine positive or negative  sin Draw diagram to determine positive or negative Find reference angle Reference angle of 210 = 210  180 = 30 Find reference angle Solve sin 210 =  sin 30 = 1 2  Solve b. Evaluate cos 150 c. Evaluate sec 135 Draw diagram to determine positive or negative Draw diagram to determine positive or negative Find reference angle Find reference angle Solve Solve 60 60 60 2 2 2 60 30 2 1 2 2 2 1 3   1 1 2 2 1 1 2   45 45 1 1 1 1 y x180 360 cos ( ) = cos  sin ( ) =  sin  tan ( ) =  tan  y x O  
  8. 8. zefry@sas.edu.my 8 5.2.4 Solve trigonometric equations A. Steps to solve trigonometric equation 1. Determine the range of the angle. 2. Find the reference angle using tables or calculator. 3. Determine the quadrant where the angle of the trigonometric function is placed. 4. Determine the values of angles in the respective quadrants. 1. Solve the following equation for 0    360. Example: sin  = 0.6428 a. cos  = 0.3420 Range : 0    360 0    360 Reference angle :  = sin1 0.6428  = 40 Quadrant : Quadrant I Quadrant II Quadrant ____ Quadrant ____ Actual angles  = 40 ,  = 180  40  = 40 , 140 b. tan  = 1.192 c. cos  =  0.7660 Range : Reference angle : Quadrant : Quadrant ___ Quadrant ___ Quadrant ___ Quadrant ___ Actual angles y x y x y x180 360 S A T C y x180 360 S A T C y x y x y x180 360 40 S A T C y x180 360 40 S A T C
  9. 9. zefry@sas.edu.my 9 d. sin  =  0.9397 e. tan  =  0.3640 Range : Reference angle : Quadrant : Quadrant ___ Quadrant ___ Quadrant ___ Quadrant ___ Actual angles f. cot  =  1.4826 g. cosec  =  2.2027 Range : Reference angle : Quadrant : Quadrant ___ Quadrant ___ Quadrant ___ Quadrant ___ Actual angles 2. Solve the following equation for 0    360. example : sec 2 = 2 a. 2 sin 2 = 1.6248 Range : 0    360 0  2  720 Reference angle : 1 2 2 1 2 2 2 60 cos cos        Quadrant : Actual angles 2 = 60, 360  60, 60 + 360, (36060) + 360  = 60, 300, 420, 660 y x y x y x y x y x y x y x y x y x180 360,720 S A T C 60 60
  10. 10. zefry@sas.edu.my 10 b. cos 3 =  0.9781 c. tan 2  =  2.05 Range Reference angle : Quadrant : Actual angles d. sin ( + 10) = 0.7660 e. cos ( + 40) = 0.7071 f. tan ( + 15) = 1 g. cos (  20) = 0.5 h. tan (2  10) =  2.082 i. sin (2  30) = 0.5
  11. 11. zefry@sas.edu.my 11 j. sin  = cos 20 k. cos  =  sin 55 Example : 2 sin x cos x = cos x 2 sin x cos x  cos x = 0 cos x ( 2 sin x  1) = 0 cos x = 0 , 2 sin x  1 = 0 sin x = 1 2 x = 30 x = 90 , 270 x = 30, 150  x = 30, 90, 150, 270 m. 2sin x cos x = sin x n. 2 cos 2  + 3 cos  =  1 o. 2 sin2  + 5 sin  = 3 p. tan2  = tan  q. 3 sin  = 2 + cosec  y x360 y x180 360 S A T C
  12. 12. zefry@sas.edu.my 12 3. Given that px sin and 00 < x < 900 . Express each of the following trigonometric ratios in terms of p. (a) sec x = (b) cosec x = (c) tan x = (d) cot x = (e) sin ( 900 - x) = (f) cos (900 - x) = (g) sec (900 - x) = (h) cosec (900 – x) = (i) tan ( 90o - x) = (j) cot ( 90o – x ) = (k) sin(-x) = (l) cos (-x) = x
  13. 13. zefry@sas.edu.my 13 4. Given that 17 8 sin x and 2700 < x < 3600 . Without using tables or calculator, find the values of. 5. Given that 17 8 -cos x and 1800 < x < 2700 . Without using tables or calculator, find the values of (a) cos x = (a) sin x = (b) tan x = (b) tan x = (c) cosec x = (c) cosec x = (d) sec x = (d) sec x = (e) cos (900 – x) = (e) sec (900 – x) = (f) sin ( 900 – x ) = (f) cot ( 900 – x ) = (g) sin (-x) = (g) sin (-x) = (h) tan (-x) = (h) cos (-x) = x x
  14. 14. zefry@sas.edu.my 14 5.4 Basic Identities 5.4.1 Prove Trigonometric Identities using Basic Identities Three basic trigonometric identities : sin 2  + cos 2  = 1 1 + tan 2  = sec 2  1 + cot 2  = cosec 2  Formula of compound angle : sin (A  B) = sin A cos B  cos A sin B cos (A  B) = cos A cos B Ŧ sin A sin B tan (A  B) = tan tan 1 tan tan A B A B  Formula of double angle : sin 2A = 2 sin A cos A cos 2A = cos2 A − sin2 A = 2 cos2 A − 1 = 1 − 2sin2 A tan 2A = A A 2 tan1 tan2  Formula of half angle : sin A = 2 sin 2 A cos 2 A cos A = cos2 2 A − sin2 2 A = 2 kos2 2 A − 1 = 1 − 2sin2 2 A tan 2A = 2 2 2 1 2 A tan A tan 1. Prove the following identities Example: cot  + tan  = cosec  sec  2 2 1 cos sin cot tan sin cos cos sin sin cos sin cos cosec sec                     a. tan2  (1  sin2 ) = sin2  b. 2 1 1 sin cos cos       c. sin2  + cot2  = cosec2   cos2  cos2  = 1 – sin2  sin2  = 1 – cos2 
  15. 15. zefry@sas.edu.my 15 d.  2 1 1 sin sec tan sin         e. 1 2 1 sin x cos x sec x cos x sin x     2. Solve the following equations for 0  x  360. a. 3 sin x + 2 = cosec x b. 2 cot2 x  5 cot x + 2 = 0 c. cos2 x  3 sin2 x + 3 = 0 d. cot2 x= 1 + cosec x e. 2 tan 2 x = 4 + sec x
  16. 16. zefry@sas.edu.my 16 ANSWERS 5.2.1 5. cos 51= 0.6293 5.2.4 2a. 0  2  720 , 54.33  = 27.17, 62.83, 207.17, 242.83 3(a) 2 1 1 p 5.(a)  17 15 6. sin 70= 0.9397 b. 0  3  1080 , 12.01  = 56, 64, 176, 184, 296, 304 (b) p 1 (b) 15 8 7. tan 25= 0.4663 c. 0  2   180 , 64  = 232 (c) 2 1 p p  (c)  17 15 5.2.2 7. sin 48 = 0.7431 d.  = 40, 120 (d) p p2 1 (d)  17 8 8. tan 67 = 2.3559 e.  = 5, 275 (e) 2 1 p (e)  17 15 9. cosec 37 = 1.6617 f.  = 30, 210 (f) p (f) 8 15 5.2.3 2a. tan 300 = 3 g.  = 80, 320 (g) p 1 (g) 15 17 b. cos 150 = 3 2  h.  = 62.83, 152.83, 242.83, 332.83 (h) 2 1 1 p (h) 17 8  c. sec 135 = 1 2  i.  = 30, 90, 210, 270 (i) p p2 1 5.2.4 1a.70 , Quadrant I, IV  = 70, 290 j.  = 70, 110 (j) 2 1 p p  b. 0    360 , 50 , Quadrant I, IV  = 30, 330 k.  = 145, 215 (k) -p c. 0    360 , 40 , Quadrant II, III  = 140, 220 m.  = 60, 180, 300 (l) 2 1 p d. 0    360 , 70 Quadrant III, IV  = 250, 290 n.  = 120, 180, 240 4.(a) 17 15 e. 0    360 , 20.01 Quadrant II, IV  = 159.99, 339.99 o.  = 30, 150 (b)  15 8 f. 0    360 , 34 Quadrant II, IV  = 146, 326 p.  = 0, 45, 225 (c)  17 8 g. 0    360 , 27 Quadrant III, IV  = 207, 333 q.  = 90, 199.47, 350.53 (d) 15 17 (e) 17 8  (f) 17 15 (g) 17 8 (h) 8 15
  17. 17. zefry@sas.edu.my 17

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