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Mathematics Question on Inverse Trigonometric Functions

sin⁡2 θ cos⁡3 θ−sin⁡4 θ cos⁡ θ\sin^{2}\,\theta\, \cos^{3}\,\theta-\sin^{4}\,\theta\, \cos\,\theta is equal

A

12cos⁡ θ sin⁡ 4θ\frac{1}{2} \cos\,\theta\, \sin\,4\theta

B

14cos⁡ θ sin⁡ 4θ\frac{1}{4} \cos\,\theta\, \sin\,4\theta

C

12sin⁡2 4θ\frac{1}{2} \sin^2\,4\theta

D

14sin⁡ θ sin⁡ 4θ\frac{1}{4} \sin\,\theta\, \sin\,4\theta

Answer

14sin⁡ θ sin⁡ 4θ\frac{1}{4} \sin\,\theta\, \sin\,4\theta

Explanation

Solution

sin⁡2θcos⁡3θ−sin⁡4θcos⁡θ\sin ^{2} \theta \cos ^{3} \theta-\sin ^{4} \theta \cos \theta =sin⁡2θcos⁡θ(cos⁡2θ−sin⁡2θ)=\sin ^{2} \theta \cos \theta\left(\cos ^{2} \theta-\sin ^{2} \theta\right) =sin⁡2θcos⁡θ(cos⁡2θ)=\sin ^{2} \theta \cos \theta(\cos 2 \theta) =sin⁡θsin⁡θcos⁡θ(cos⁡2θ)=\sin \theta \sin \theta \cos \theta(\cos 2 \theta) =sin⁡θ⋅22sin⁡θcos⁡θ(cos⁡2θ)=\sin \theta \cdot \frac{2}{2} \sin \theta \cos \theta(\cos 2 \theta) =12sin⁡θsin⁡2θcos⁡2θ=\frac{1}{2} \sin \theta \sin 2 \theta \cos 2 \theta [∵2sin⁡θcos⁡θ=sin⁡2θ][\because 2 \sin \theta \cos \theta=\sin 2 \theta] =12sin⁡θ⋅22sin⁡2θcos⁡2θ=14sin⁡θ⋅sin⁡4θ=\frac{1}{2} \sin \theta \cdot \frac{2}{2} \sin 2 \theta \cos 2 \theta=\frac{1}{4} \sin \theta \cdot \sin 4 \theta