Question

Prove that expression $\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{\sin 16x}{16\sin x}$

Answer 1

Hint: In this question, you can use sine double angle formula. The sine double angle formula tells us that $\sin 2\theta $ is always equal to $2\sin \theta \cos \theta $. You can use this identity to rewrite expressions and prove.

Complete step-by-step answer:
Let us consider the L. H. S. $\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x$ and multiply and divide by the trigonometric sine function $\sin x$.
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{\sin x\cdot \cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x}{\sin x}\]
Now multiply and divided by 2, we get
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{2\sin x\cdot \cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x}{2\sin x}\]
The sine double angle formula tells us that $2\sin x\cos x$ is always equal to $\sin 2x$.
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{\sin 2x\cdot cox2x\cdot \cos 4x\cdot \cos 8x}{2\sin x}\]
Again, multiply and divided by 2, we get
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{2\sin 2x\cdot cox2x\cdot \cos 4x\cdot \cos 8x}{2\times 2\times \sin x}\]
The sine double angle formula tells us that $2\sin 2x\cos 2x$ is always equal to $\sin 4x$.
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{\sin 4x\cdot \cos 4x\cdot \cos 8x}{2\times 2\times \sin x}\]
Also, multiply and divided by 2, we get
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{2\sin 4x\cdot \cos 4x\cdot \cos 8x}{2\times 2\times 2\times \sin x}\]
The sine double angle formula tells us that $2\sin 4x\cos 4x$ is always equal to $\sin 8x$.
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{\sin 8x\cdot \cos 8x}{2\times 2\times 2\times \sin x}\]
Again, multiply and divided by 2, we get
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{2\sin 8x\cdot \cos 8x}{2\times 2\times 2\times 2\times \sin x}\]
The sine double angle formula tells us that $2\sin 8x\cos 8x$ is always equal to $\sin 16x$.
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{\sin 16x}{2\times 2\times 2\times 2\times \sin x}\]
\[\cos x\cdot cox2x\cdot \cos 4x\cdot \cos 8x=\dfrac{\sin 16x}{16\sin x}\]
Hence proved.

Note: Alternatively, The given question is solved as follows. Let us consider $\sin (16x)$ and apply the sine double angle formula that tells us $\sin (16x)$ is always equal to $2\sin 8x\cdot \cos 8x$.
 $\sin (16x)=2\sin 8x\cos 8x$
The sine double angle formula tells us that $\sin 8x$ is always equal to $2\sin 4x\cos 4x$.
$\sin (16x)=2(2\sin 4x\cos 4x)\cos 8x$
$\sin (16x)=4\sin 4x\cos 4x\cos 8x$
The sine double angle formula tells us that $\sin 4x$ is always equal to $2\sin 2x\cos 2x$.
$\sin (16x)=4(2\sin 2x\cos 2x)\cos 4x\cos 8x$
$\sin (16x)=8\sin 2x\cos 2x\cos 4x\cos 8x$
The sine double angle formula tells us that $\sin 2x$ is always equal to $2\sin x\cos x$.
$\sin (16x)=8(2\sin x\cos x)\cos 2x\cos 4x\cos 8x$
$\sin (16x)=16\sin x\cos x\cos 2x\cos 4x\cos 8x$
Dividing both sides by $16\sin x$, we get
 $\dfrac{\sin (16x)}{16\sin x}=\cos x\cos 2x\cos 4x\cos 8x$
Hence proved.