Prove that =32 cos3- 32cos3+6cos
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Step-by-step explanation:
sin6θ=2sin3θcos3θ
=2(3sinθ−4sin
3
θ)(4cos
3
θ−3cosθ)
=24sinθcos
3
θ−18sinθcosθ−32sin
3
θcos
3
θ+24sin
3
θcosθ
=24sinθcosθ(1−sin
2
θ)−18sinθcosθ−32sinθ(1−cos
2
θ)cos
3
θ+24sin
3
θcosθ
=24sinθcosθ−24sin
3
θcosθ−18sinθcosθ−32sinθcos
3
θ+32cos
5
θsinθ+24sin
3
θcosθ
=32cos
5
θsinθ−32sinθcos
3
θ+6sinθcosθ
=32cos
5
θsinθ−32sinθcos
3
θ+3sin2θ
Therefore x=sin2θ
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