Question

Prove the following trigonometric equation:
$\sin x+\sin 3x+\sin 5x+\sin 7x=4\cos x\cos 2x\sin 4x$

Answer 1

Hint:To prove the above problem, we will simplify the L.H.S of the given equation. In the L.H.S, you can see the 4 sine terms so we can apply the identity of $\sin C+\sin D$ in these 4 terms taking the sine terms as two at a time then simplify.

Complete step-by-step answer:
The equation that we have to prove is:
$\sin x+\sin 3x+\sin 5x+\sin 7x=4\cos x\cos 2x\sin 4x$
We are going to simplify the L.H.S of the above equation and try to resolve into R.H.S.
$\sin x+\sin 3x+\sin 5x+\sin 7x$
As you can see the trigonometric expressions are in the form of $\sin C+\sin D$ taken two at a time. We are going to show below how the above sine terms are in the form of $\sin C+\sin D$.
$\begin{align}
  & \sin 7x+\sin x=\sin C+\sin D \\
 & \sin 5x+\sin 3x=\sin C+\sin D \\
\end{align}$
Now, we know the identity of $\sin C+\sin D$:
$\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
Applying the above identity in $\sin x+\sin 3x+\sin 5x+\sin 7x$ we get,
$\begin{align}
  & 2\sin \left( \dfrac{7x+x}{2} \right)\cos \left( \dfrac{7x-x}{2} \right)+2\sin \left( \dfrac{5x+3x}{2} \right)\cos \left( \dfrac{5x-3x}{2} \right) \\
 & =2\sin 4x\cos 3x+2\sin 4x\cos x \\
\end{align}$
Now, taking $2\sin 4x$ as common from the above expression we get,
$2\sin 4x\left( \cos 3x+\cos x \right)$
As you can see from the above expression that $\cos 3x+\cos x$ is in the form of $\cos C+\cos D$ so we are going to apply the identity of $\cos C+\cos D$ on $\cos 3x+\cos x$.
We know that the identity of $\cos C+\cos D$ is:
$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
Applying the above identity in $2\sin 4x\left( \cos 3x+\cos x \right)$ we get,
$\begin{align}
  & 2\sin 4x\left( 2\cos \left( \dfrac{3x+x}{2} \right)\cos \left( \dfrac{3x-x}{2} \right) \right) \\
 & =2\sin 4x\left( 2\cos 2x\cos x \right) \\
 & =4\sin 4x\cos 2x\cos x \\
\end{align}$
The R.H.S of the given equation is $4\cos x\cos 2x\sin 4x$.
From the above simplification of L.H.S of the given equation, the answer comes out to be$4\sin 4x\cos 2x\cos x$ which is equal to R.H.S of the given equation.
Hence, we have proved that L.H.S = R.H.S of the given equation.

Note: You might be thinking to apply $\sin C+\sin D$ identity on $\sin x+\sin 3x$ and $\sin 5x+\sin 7x$ instead of the one we have shown above.
Yes, you can apply the identity on $\sin x+\sin 3x$ and $\sin 5x+\sin 7x$. The answer won’t change.
Let us see how it’s not going to change.
$\sin x+\sin 3x+\sin 5x+\sin 7x=4\cos x\cos 2x\sin 4x$
Simplifying L.H.S of the above equation we get,
$\begin{align}
  & 2\sin \left( \dfrac{3x+x}{2} \right)\cos \left( \dfrac{3x-x}{2} \right)+2\sin \left( \dfrac{7x+5x}{2} \right)\cos \left( \dfrac{7x-5x}{2} \right) \\
 & =2\sin 2x\cos x+2\sin 6x\cos x \\
\end{align}$
Taking $2\cos x$ as common from the above expression we get,
$2\cos x\left( \sin 2x+\sin 6x \right)$
Applying $\sin C+\sin D$ identity on $\sin 2x+\sin 6x$ we get,
$\begin{align}
  & 2\cos x\left( 2\sin \left( \dfrac{6x+2x}{2} \right)\cos \left( \dfrac{6x-2x}{2} \right) \right) \\
 & =4\cos x\sin 4x\cos 2x \\
\end{align}$
R.H.S is equal to $4\cos x\cos 2x\sin 4x$.
From the above simplification of L.H.S, the answer comes out to be $4\cos x\sin 4x\cos 2x$ which is equal to R.H.S.