SIR PLEASE PROVE

Asked by gunjansingh | 9th Dec, 2009, 10:08: PM

Expert Answer:

sinθ + cos θ = x 

Squaring both sides we get

sin2θ + cos2 θ +2sin θcos θ = x 2

sin θcos θ =(x 2 -1)/2

sin2θ + cos2 θ  =1

cubing both sides we have

sin 6 θ+ cos 6 θ + 3sin θcos θ (sin2θ + cos2 θ)= 1

sin 6 θ+ cos 6 θ  = 1- 3(x 2 -1)2/4  = (4- 3((x 2 -1)2/4)

 

Answered by  | 10th Dec, 2009, 11:16: AM

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