Prove that: 5 x=5 x-20 ^3 x+16 ^5 x.

Question

Prove that: $\sin 5 x=5 \sin x-20 \sin ^3 x+16 \sin ^5 x$.

✓

Answer

We have to prove that $\sin 5 x=5 \sin x-20 \sin ^3 x+16 \sin ^5 x$.
Let us consider $\text{LHS} = \sin 5x$
$\sin 5x = \sin(3x + 2x)$
But we know,
$\sin(x + y) = \sin x \cos y + \cos x \sin y ... (i)$
$\Rightarrow \sin 5 x=\sin 3 x \cos 2 x+\cos 3 x \sin 2 x$
$\Rightarrow \sin 5 x=\sin (2 x+x) \cos 2 x+\cos (2 x+x) \sin 2 x \ldots(ii)$
and
$\cos (x + y) = \cos(x)\cos(y) - \sin(x)\sin(y) ... (iii)$
Now substituting equation $(i)$ and $(iii)$ in equation $(ii),$ we get
$\Rightarrow \sin 5 x=(\sin 2 x \cos x +\cos 2 x \sin x ) \cos 2 x +(\cos 2 x \cos x -\sin 2 x \sin x ) \sin 2 x$
$\Rightarrow \sin 5 x =\sin 2 x$
$\Rightarrow \sin 5 x +\left(\sin 2 x \cos 2 x \cos x -\sin ^2 2 x \sin 2 x \right)$
Now $\sin 2 x =2 x \sin x \cos x +\cos ^2 2 x \sin x -\sin ^2 2 x \sin x \ldots \text { (iv) }$
And $\cos 2 x =\cos ^2 x -\sin ^2 x \ldots(vi)$
Substituting equation $(v)$ and $(vi)$ in equation $(iv),$ we get
$\Rightarrow \sin 5 x=2(2 \sin x \cos x)\left(\cos ^2 x-\sin ^2 x\right) \cos x+\left(\cos ^2 x-\sin ^2 x\right)^2 \sin x-(2 \sin x \cos x)^2 \sin x$
$\Rightarrow \sin 5 x=4\left(\sin x \cos ^2 x\right)\left(\left[1-\sin ^2 x\right]-\sin ^2 x\right)+\left(\left[1-\sin ^2 x\right]-\sin ^2 x\right)^2 \sin _x x$
$-\left(4 \sin ^2 x \cos ^2 x\right) \sin _x(\operatorname{as} \cos ^2 x+\sin ^2 x=1$
$\Rightarrow \cos ^2 x=.1-\sin ^2 x)$
$\Rightarrow \sin 5 x=4\left(\sin x\left[1-\sin ^2 x\right]\right)\left(1-2 \sin ^2 x\right)+\left(1-2 \sin ^2 x\right)^2 \sin x-4 \sin ^3 x\left[1-\sin ^2 x\right]$
$\Rightarrow \sin 5 x=4 \sin x\left(1-\sin ^2 x\right)\left(1-2 \sin ^2 x\right)+\left(1-4 \sin ^2 x+4 \sin ^4 x\right) \sin x-4 \sin ^3 x+4 \sin ^5 x$
$\Rightarrow \sin 5 x=\left(4 \sin x-4 \sin ^3 x\right)\left(1-2 \sin ^2 x\right)+\sin x-4 \sin ^3 x+4 \sin ^5 x-4 \sin ^3 x+4 \sin ^5 x$
$\Rightarrow \sin 5 x=4 \sin x-8 \sin ^3 x-4 \sin ^3 x+8 \sin ^5 x+\sin x-8 \sin ^3 x+8 \sin ^5 x$
$\Rightarrow \sin 5 x=5 \sin x-20 \sin ^3 x+16 \sin ^5 x$
Hence $\text{LHS = RHS}$
Hence proved.

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