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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