Question

How do you prove \[\sin 8x=8\sin x\cos x\cos 2x\cos 4x\]?

Answer 1

Hint: We solve this question using basic trigonometric identities. We can solve the equation from both LHS and RHS. Here we will start form LHS and by using the formula of \[\sin 2x\]we will derive the equation repeat the step by substituting the required formula until we arrive at the solution.

Complete step by step answer:
First we know the formulas that are required to solve the problem.
\[\sin 2x=2\sin x\cos x\]
\[\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\]
\[\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}\]
Here in this problem we use only \[\sin 2x\] for derivation. But we should use all the formulas accordingly to get the required solution given in the RHS side.
To solve the given equation
We will start with the LHS \[\sin 8x\]
We can write
\[\sin 8x=\sin 2\left( 4x \right)\]
So we have formula for \[\sin 2x\] by deriving our equation using this formula we get
\[\sin 2\left( 4x \right)=2\sin 4x\cos 4x\]
Again we have \[2\sin 4x\cos 4x\]
We can write the equation again as
\[\Rightarrow 2\sin 2\left( 2x \right)\cos 4x\]
Again we have to substitute the formula of \[\sin 2x\] in the above equation
\[2\sin 2\left( 2x \right)\cos 4x=4\sin 2x\cos 2x\cos 4x\]
Now again we have \[\sin 2x\] in the equation so again we have to substitute the formula
\[4\sin 2x\cos 2x\cos 4x=8\sin x\cos x\cos 2x\cos 4x\]
From this we can conclude that
\[4\sin 2x\cos 2x\cos 4x=8\sin x\cos x\cos 2x\cos 4x\]

Note:
We can do this from the RHS side also. Here we had substituted the required formula continuously until we arrived at the RHS equation. Like this we can do it in reverse direction from RHS to derive the LHS. In this process we are deriving the LHS. In another approach using RHS we solve the problem by merging into a formula from the derived one. But we should be aware of formulas which we need to apply to arrive at the correct solution.