Question

Prove the following expression
$\sin x+\sin 3x+\sin 5x+\sin 7x=4\cos x\cos 2x\sin 4x$

Answer 1

Hint: To solve this question, recall all the formulas that you have studied in trigonometry. There is a formula in trigonometry for addition of two $\sin$ functions with different arguments. The formula is $\sin A+\sin B=2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)$. Use this formula to solve this question.

Complete step-by-step answer:

Before proceeding with the question, one must know all the formulas that will be required to solve this question. In trigonometry, we have a formula that can be applied to the sum of two $\sin$ functions with different arguments. That formula is,
$\sin A+\sin B=2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)..............\left(1\right)$
Also, in trigonometry, there is a formula for $\cos$ function. That formula is,
$\cos (-A)=\cos \left(A\right)..............\left(2\right)$
In this question, we have to prove $\sin x+\sin 3x+\sin 5x+\sin 7x=4\cos x\cos 2x\sin 4x$.
Let us start from the left side of the expression. On the left side, we have,
$$\sin x+\sin 3x+\sin 5x+\sin 7x$$
We have to prove this equal to the right side of the expression i.e. $4\cos x\cos 2x\sin 4x$.
Let us club some terms on the left side with the use of brackets.
$(\sin x+\sin 3x)+(\sin 5x+\sin 7x)....................\left(3\right)$
Inside the individual bracket, we can see that the term is the sum of two $\sin$ functions with different arguments. Hence, we can apply formula $\left(1\right)$ in both the brackets.
Applying formula $\left(1\right)$ to the first bracket i.e. $\sin x+\sin 3x$, we get,
$$\begin{array}{rl}& \sin x+\sin 3x=2\sin \left({\displaystyle \frac{x+3x}{2}}\right)\cos \left({\displaystyle \frac{x-3x}{2}}\right)\\ & \Rightarrow \sin x+\sin 3x=2\sin \left({\displaystyle \frac{4x}{2}}\right)\cos \left({\displaystyle \frac{-2x}{2}}\right)\\ & \Rightarrow \sin x+\sin 3x=2\sin \left(2x\right)\cos (-x)\end{array}$$
From the formula $\left(2\right)$, we can say $\cos (-x)=\cos x$.
$$\Rightarrow \sin x+\sin 3x=2\sin 2x\cos x............\left(4\right)$$
Applying formula $\left(1\right)$ to the second bracket i.e. $\sin 5x+\sin 7x$, we get,
$$\begin{array}{rl}& \sin 5x+\sin 7x=2\sin \left({\displaystyle \frac{5x+7x}{2}}\right)\cos \left({\displaystyle \frac{5x-7x}{2}}\right)\\ & \Rightarrow \sin 5x+\sin 7x=2\sin \left({\displaystyle \frac{12x}{2}}\right)\cos \left({\displaystyle \frac{-2x}{2}}\right)\\ & \Rightarrow \sin 5+\sin 7x=2\sin \left(6x\right)\cos (-x)\end{array}$$
From the formula $\left(2\right)$, we can say $\cos (-x)=\cos x$.
$$\Rightarrow \sin 5x+\sin 7x=2\sin 6x\cos x............\left(5\right)$$
Substituting equation $\left(4\right)$ and equation $\left(5\right)$ in the expression $\left(3\right)$, we get,
 $\begin{array}{rl}& (\sin x+\sin 3x)+(\sin 5x+\sin 7x)=2\sin 2x\cos x+2\sin 6x\cos x\\ & \Rightarrow (\sin x+\sin 3x)+(\sin 5x+\sin 7x)=2\cos x(\sin 2x+\sin 6x)..................\left(6\right)\end{array}$
Applying formula $\left(1\right)$ to $\sin 2x+\sin 6x$ in the above expression, we get,
$\begin{array}{rl}& \sin 2x+\sin 6x=2\sin \left({\displaystyle \frac{2x+6x}{2}}\right)\cos \left({\displaystyle \frac{2x-6x}{2}}\right)\\ & \Rightarrow \sin 2x+\sin 6x=2\sin \left({\displaystyle \frac{8x}{2}}\right)\cos \left({\displaystyle \frac{-4x}{2}}\right)\\ & \Rightarrow \sin 2x+\sin 6x=2\sin \left(4x\right)\cos (-2x)\end{array}$
From the formula $\left(2\right)$, we can say $\cos (-2x)=\cos 2x$.
$\Rightarrow \sin 2x+\sin 6x=2\sin 4x\cos 2x$
Substituting $\sin 2x+\sin 6x=2\sin 4x\cos 2x$ in equation$\left(6\right)$, we get,
$\begin{array}{rl}& (\sin x+\sin 3x)+(\sin 5x+\sin 7x)=2\cos x(2\sin 4x\cos 2x)\\ & \Rightarrow (\sin x+\sin 3x)+(\sin 5x+\sin 7x)=4\cos x\cos 2x\sin 4x\end{array}$
Hence, we have proved the left side of the expression in the question to it’s right side.

Note: One can also solve this question by simplifying the right side term using the formula $2\cos a\cos b=\cos (a+b)+\cos (a-b)$ and $2\sin a\cos b=\sin (a+b)+\sin (a-b)$. Simplifying the right side using these formulas, we will get the left side.