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

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 LHS = sin 5x
sin 5x = sin(3x + 2x)
But we know,
sin(x + y) = sin x cos y + cos x sin y ... (i)
$\begin{array}{l}\Rightarrow \sin 5 x=\sin 3 x \cos 2 x+\cos 3 x \sin 2 x \\ \Rightarrow \sin 5 x=\sin (2 x+x) \cos 2 x+\cos (2 x+x) \sin 2 x \ldots \text { (ii) }\end{array}$
and
cos (x + y) = cos(x)cos(y) - sin(x)sin(y) ... (iii)
Now substituting equation (i) and (iii) in equation (ii), we get
$\begin{array}{l}\Rightarrow \sin 5 x=(\sin 2 x \cos x +\cos 2 x \sin x ) \cos 2 x +(\cos 2 x \cos x -\sin 2 x \sin x ) \sin 2 x \\ \Rightarrow \sin 5 x =\sin 2 x c cop text \\ \Rightarrow \sin 5 x +\left(\sin 2 x \cos 2 x \cos x -\sin ^2 2 x \sin 2 x \right) \\ \text { Now } \sin 2 x =2 x \sin x \cos x +\cos ^2 2 x \sin x -\sin ^2 2 x \sin x \ldots \text { (iv) } \\ \text { And } \cos 2 x =\cos ^2 x -\sin ^2 x \ldots \text { (vi) }\end{array}$
Substituting equation (v) and (vi) in equation (iv), we get
$\begin{array}{l}\Rightarrow \sin 5 x=2(2 \sin x \cos x)\left(\cos ^2 x-\sin ^2 x\right) \cos x+\left(\cos ^2 x-\sin ^2 x\right)^2 \sin x-(2 \sin x \cos x)^2 \sin x \\ \Rightarrow \sin 5 x=4\left(\sin x \cos ^2 x\right)\left(\left[1-\sin ^2 x\right]-\sin ^2 x\right)+\left(\left[1-\sin ^2 x\right]-\sin ^2 x\right)^2 \sin _x x \\ -\left(4 \sin ^2 x \cos ^2 x\right) \sin _x\left(\operatorname{as} \cos ^2 x+\sin ^2 x=1 \Rightarrow \cos ^2 x=\right. \ \left.1-\sin ^2 x\right) \\ \Rightarrow \sin 5 x=4\left(\sin x\left[1-\sin ^2 x\right]\right)\left(1-2 \sin ^2 x\right)+\left(1-2 \sin ^2 x\right)^2 \sin x-4 \sin ^3 x\left[1-\sin ^2 x\right] \\ \Rightarrow \sin 5 x=4 \sin x\left(1-\sin ^2 x\right)\left(1-2 \sin ^2 x\right)+\left(1-4 \sin ^2 x+4 \sin ^4 x\right) \sin x-4 \sin ^3 x+4 \sin ^5 x \\ \Rightarrow \sin 5 x=\left(4 \sin x-4 \sin ^3 x\right)\left(1-2 \sin ^2 x\right)+\sin x-4 \sin ^3 x+4 \sin ^5 x-4 \sin ^3 x+4 \sin ^5 x \\ \Rightarrow \sin 5 x=4 \sin x-8 \sin ^3 x-4 \sin ^3 x+8 \sin ^5 x+\sin x-8 \sin ^3 x+8 \sin ^5 x \\ \Rightarrow \sin 5 x=5 \sin x-20 \sin ^3 x+16 \sin ^5 x\end{array}$
Hence LHS = RHS
Hence proved.

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