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Trigo functions

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• 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. 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. 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. 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. 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  