1 / 23

Trigonometric Identities

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

allie
Download Presentation

Trigonometric Identities

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Trigonometric Identities

  2. 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.

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

  4. Examples R.H.S. = R.H.S.

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. Examples L.H.S. = R.H.S.

  13. 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

  14. Double angle formulae

  15. 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

  16. 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 = 

  17. 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 

  18. 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

  19. 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 +  )

  20. 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 )

  21. 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 )

  22. 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 )

  23. 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

More Related