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Reciprocal Trigonometry Functions

Reciprocal Trigonometry Functions. Cosecant,. Secant. and Cotangent. Provided sin x  0, cos x  0 and tan x  0. Third letter rule. Example: Find (3 dps). Answers:. (i) 1.035 (ii) -3.236 (iii) -0.176. Cosec x. x. Graphs of cosec, sec and cot.

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Reciprocal Trigonometry Functions

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  1. Reciprocal Trigonometry Functions Cosecant, Secant and Cotangent Provided sin x 0, cos x  0 and tan x  0 Third letter rule Example: Find (3 dps) Answers: (i) 1.035 (ii) -3.236 (iii) -0.176

  2. Cosec x x Graphs of cosec, sec and cot The graphs of the reciprocal functions can be found by taking the corresponding sine, cosine and tangent graph and calculating the reciprocals of each point on the graph.

  3. sec x x Graphs of cosec, sec and cot

  4. cot x x Graphs of cosec, sec and cot

  5. A S C T Examples Find the exact values of: Answers

  6. A S C T Examples Given that sin A =4/5, where A is obtuse, and cosB = 3/2, where B is acute, find the exact values of: Answers

  7. Trigonometric Identities

  8. Examples Prove that (1 – cos A)(1 + sec A)  sin A tan A L.H.S. (1 – cos A)(1 + sec A) = 1 + sec A – cos A –Cos A sec A = 1 + sec A – cos A - 1 = sec A – cos A = sin A tan A = R.H.S.

  9. Examples Prove that cot A + tan A  sec A cosec A L.H.S. = R.H.S.

  10. Examples R.H.S. = R.H.S.

  11. Solving equations Solve 2 tan2 x – 7 sec x + 8 = 0 for 0  x  360 2 (sec2x – 1) – 7 sec x + 8 = 0 2 sec2x – 2 – 7 sec x + 8 = 0 2 sec2x – 7 sec x + 6 = 0 (2 sec x – 3)(sec x – 2)= 0 sec x = 3/2 or sec x = 2 cos x = 2/3 or cos x = ½ x = 48.2 or x = 60 or: x = 360 – 48.2 or x = 360 - 60 complete solution: x = 48.2 or 60 or 300 or 311.8

  12. Solving equations Solve 2 cos x = cot x for 0  x  360 2 cos x = cos x/ sin x 2 cos x sin x = cos x 2 cos x sin x – cos x= 0 cos x(2 sin x – 1)= 0 cos x = 0 or sin x = ½ cos x = 0 x = 90 or 270 sin x = ½  x = 30 or 330 complete solution: x = 30 or 90 or 270 or 30

  13. Solving equations Solve 3 cot2 x – 10 cot x + 3 = 0 for 0  x  2 (3 cot x - 1)(cot x – 3) = 0 cot x = 1/3 or cot x = 3  tan x = 3 or tan x = 1/3 tan x = 3 x = 1.24c or 4.39c tan x = 1/3  x = 0.32c or 3.46c complete solution: x = 0.32c or 1.24c or 3.46c or 4.39c

  14. Solving equations Solve 5 cot2 x – 2 cosec x + 2 = 0 for 0  x  2 5(cosec2 x – 1) – 2 cosec x + 2 = 0 5cosec2 x – 5 – 2 cosec x + 2 = 0 5cosec2 x – 2 cosec x - 3 = 0 sin x = -5/3 not possible or sin x = 1  x = /2

  15. Additional formulae sin (A + B) = sin A cos B +sin B cos A sin (A - B) = sin A cos B -sin B cos A cos (A + B) = cos A cos B- sin A sin B cos (A - B) = cos A cos B+ sin A sin B

  16. Examples Find the exact value of sin 75 sin (A + B) = sin A cos B +sin B cos A sin (30 + 45) = sin 30 cos 45 +sin 45 cos 30

  17. Examples Express cos (x + /3) in terms of cos x and sin x cos (A + B) = cos A cos B- sin A sin B cos (x + /3) = cos x cos /3- sin /3sin x

  18. Examples L.H.S. = R.H.S.

  19. Double angle formulae sin (A + B) = sin A cos B +sin B cos A sin (A + A) = sin A cos A + sin A cos A sin 2A = 2 sin A cos A cos (A + B) = cos A cos B - sin A sin B cos (A + A) = cos A cos A- sin A sin A cos (A + A) = cos2A - sin2A cos 2A = cos2A - sin2A cos 2A = 2cos2A - 1 cos 2A = 1 – 2sin2A

  20. Double angle formulae

  21. 4 1 A 15 Examples Given that cos A = 2/3, find the exact value of cos 2A. cos 2A = 2cos2A - 1 Given that sin A = ¼ , find the exact value of sin 2A. sin 2A = 2 sin A cos A

  22. Solving equations Solve cos 2A + 3 + 4 cos A = 0 for 0  x  2 =2 cos2A - 1+ 3 + 4 cos A = 0 =2 cos2A + 4 cos A + 2= 0 = cos2A + 2 cos A + 1 = 0 = cos2A + 2 cos A + 1 = 0 = (cos A + 1)2 = 0 = cos A = - 1  A = 

  23. Solving equations Solve sin 2A = sin A for -  x   =2sin A cos A = sin A =2 sin A cos A – sin A = 0 = sin A(2 cos A – 1) = 0 sin A = 0 or cos A = ½ sin A = 0  A = -  or 0 or  cos A = ½  A = - /3 or /3 Complete solution: A = -  or - /3 or 0 or /3 or 

  24. Solving equations Solve tan 2A + 5 tan A = 0 for 0 x  2 tan A = 0  A = 0 or  or 2 7 – 5tan2 A = 0 tan A =  7/5  A = 0.97 , 2.27, 4.01 or 5.41c Complete solution: A= 0.97 , 2.27, 4.01, 5.41c 0,  or 2

  25. Harmonic form If a and b are positive a sin x + b cos x can be written in the form R sin( x +  ) a sin x - b cos x can be written in the form R sin( x -  ) a cos x + b sin x can be written in the form R cos( x -  ) a cos x - b sin x can be written in the form R cos( x +  )

  26. Examples Express 3 cos x + 4 sin x in the form R cos( x -  ) R cos( x -  ) = R cos x cos + R sin x sin  3 cos x + 4 sin x= R cos x cos + R sin x sin  R cos  = 3 [1] R sin  = 4 [2] [1]2 + [2]2 : R2 sin2 x + R2 cos2 x = 32 + 42 R2(sin2 x + cos2 x ) = 32 + 42 R2= 32 + 42 = 25  R = 5 [2]  [1]: tan  = 4/3   = 53.1 3 cos x + 4 sin x = 5 cos( x + 53.1 )

  27. Examples Express 12 cos x + 5 sin x in the form R sin( x +  ) R sin( x +  ) = R sin x cos + R cos x sin  12 cos x + 5 sin x= R sin x cos + R cos x sin  R cos  = 12 [1] R sin  = 5 [2] [1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 122 + 52 R2(cos2 x + sin2 x ) = 122 + 52 R2= 122 + 52 = 169  R = 13 [2]  [1]: tan  = 5/12   = 22.6 12 cos x + 5 sin x = 13 sin( x + 22.6 )

  28. Examples Express cos x - 3 sin x in the form R cos( x +  ) R cos( x +  ) = R cos x cos - R sin x sin  cos x - 3 sin x = R cos x cos  - R sin x sin  R cos  = 1 [1] R sin  = 3 [2] [1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 12 + (3 ) 2 R2(cos2 x + sin2 x ) = 12 + 3 R2= 1 + 3 = 4  R = 2 [2]  [1]: tan  = 3   = 60 cos x + 3 sin x = 2 cos( x + 60 )

  29. Solving equations Solve 7 sin x + 3 cos x = 6 for 0 x  2 R sin( x +  ) = R sin x cos + R cos x sin  7 sin x + 3 cos x= R sin x cos + R cos x sin  R cos  = 7 [1] R sin  = 3 [2] R2 = 72 + 32  R = 7.62 [2]  [1]: tan  = 3/7   = 0.405c (Radians) 7 sin x + 3 cos x = 7.62 sin( x + 0.405) 7.62 sin( x + 0.405 ) = 6  x + 0.405 = sin-1(6/7.62) x + 0.405 = 0.907 or 2.235 x = 0.502c or 1.830c

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