{ \sin \theta } { \cos \theta } = 32 \sin ^ { 5 } \theta - 32 \sin ^ { 3 } \theta + \sin \theta
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Answer:
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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