# Trigonometric Identities

## Presentation on theme: "Trigonometric Identities"— Presentation transcript:

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.

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

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/ or sec x = 2 cos x = 2/ or cos x = ½ x = or x = 60 or: x = 360 – or x = complete solution: x = 48.2 or 60 or 300 or 311.8

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

Solving equations Solve 3 cot2 x – 10 cot x + 3 = 0 for 0  x  2
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

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

Additional formulae 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

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

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 /3 sin x

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

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

Double angle formulae

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 A 4 1 15

Solving equations Solve cos 2A + 3 + 4 cos A = 0 for 0  x  2

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 

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

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 +  )

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  = [2] [1]2 + [2]2 : R2 sin2 x + R2 cos2 x = R2(sin2 x + cos2 x ) = R2= = 25  R = 5 [2]  [1]: tan  = 4/3   = 53.1 3 cos x + 4 sin x = 5 cos( x  )

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  = [1] R sin  = [2] [1]2 + [2]2 : R2 cos2 x + R2 sin2 x = R2(cos2 x + sin2 x ) = R2= = 169  R = 13 [2]  [1]: tan  = 5/12   = 22.6 12 cos x + 5 sin x = 13 sin( x  )

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] R sin  =  [2] [1]2 + [2]2 : R2 cos2 x + R2 sin2 x = 12 + (3 ) 2 R2(cos2 x + sin2 x ) = R2= = 4  R = 2 [2]  [1]: tan  = 3   = 60 cos x + 3 sin x = 2 cos( x + 60 )

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  = [2] R2 =  R = 7.62 [2]  [1]: tan  = 3/7   = 0.405c (Radians) 7 sin x + 3 cos x = 7.62 sin( x ) 7.62 sin( x ) = 6  x = sin-1(6/7.62) x = or x = 0.502c or c