Question

Prove the following:
$\sin \left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x=\cos x$

Answer 1

Hint: For solving this question, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side. And we will use the trigonometric formula $\cos A\cos B+\sin A\sin B=\cos \left( A-B \right)$ for simplifying the term on the left-hand side. After that, we will easily prove the desired result.

Complete step-by-step answer:
Given:
We have to prove the following equation:
$\sin \left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x=\cos x$
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
$\cos A\cos B+\sin A\sin B=\cos \left( A-B \right).....................\left( 1 \right)$
Now, we will use the above formula to simplify the term on the left-hand side.
On the left-hand side, we have $\sin \left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x$ .
Now, let $A=\left( n+2 \right)x$ and $B=\left( n+1 \right)x$ . Then,
$\begin{aligned} & \sin \left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x \\\ & \Rightarrow \sin B\sin A+\cos B\cos A \\\ & \Rightarrow \cos A\cos B+\sin A\sin B \\\ \end{aligned}$
Now, we will use the formula from equation (1) to write $\cos A\cos B+\sin A\sin B=\cos \left( A-B \right)$ in the above expression. Then,
$\begin{aligned} & \cos A\cos B+\sin A\sin B \\\ & \Rightarrow \cos \left( A-B \right) \\\ \end{aligned}$
Now, as per our assumption $A=\left( n+2 \right)x$ and $B=\left( n+1 \right)x$ . So, we can put $A=\left( n+2 \right)x$ and $B=\left( n+1 \right)x$ in the above expression. Then,
$\begin{aligned} & \cos \left( A-B \right) \\\ & \Rightarrow \cos \left( \left( n+2 \right)x-\left( n+1 \right)x \right) \\\ & \Rightarrow \cos \left( x\left( n+2-n-1 \right) \right) \\\ & \Rightarrow \cos x \\\ \end{aligned}$
Now, from the above result, we conclude that the value of the expression $\sin \left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x$ will be equal to the value of the expression $\cos x$ . Then,
$\sin \left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x=\cos x$
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, $\sin \left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x=\cos x$ .
Hence, proved.

Note: Here, the student should first understand what we have to prove in the question. After that, we should proceed in a stepwise manner and apply trigonometric formulas like $\cos A\cos B+\sin A\sin B=\cos \left( A-B \right)$ correctly. Moreover, while simplifying we should be aware of the result and avoid calculation mistakes while solving.