Question

Prove the following:
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x$

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 trigonometric formulas of $\cos \left( A+B \right)$ , $\cos \left( A-B \right)$ and $\sin \left( \pi -\theta \right)$ 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:
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x$
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:
$\begin{aligned} & \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B.............\left( 1 \right) \\\ & \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B.............\left( 2 \right) \\\ & \sin \left( \pi -\theta \right)=\sin \theta .............................................\left( 3 \right) \\\ & \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}......................................................\left( 4 \right) \\\ \end{aligned}$
Now, we will use the above four formulas to simplify the term on the left-hand side.
On the left-hand side, we have $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)$ .
Now, we will use the formula from the equation (1) to write $\cos \left( \dfrac{3\pi }{4}+x \right)=\cos \dfrac{3\pi }{4}\cos x-\sin \dfrac{3\pi }{4}\sin x$ and formula from the equation (2) to write $\cos \left( \dfrac{3\pi }{4}-x \right)=\cos \dfrac{3\pi }{4}\cos x+\sin \dfrac{3\pi }{4}\sin x$ in the term on the left-hand side. Then,
$\begin{aligned} & \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right) \\\ & \Rightarrow \left( \cos \dfrac{3\pi }{4}\cos x-\sin \dfrac{3\pi }{4}\sin x \right)-\left( \cos \dfrac{3\pi }{4}\cos x+\sin \dfrac{3\pi }{4}\sin x \right) \\\ & \Rightarrow \cos \dfrac{3\pi }{4}\cos x-\sin \dfrac{3\pi }{4}\sin x-\cos \dfrac{3\pi }{4}\cos x-\sin \dfrac{3\pi }{4}\sin x \\\ & \Rightarrow -2\sin \dfrac{3\pi }{4}\sin x \\\ \end{aligned}$
Now, we will write $\sin \dfrac{3\pi }{4}=\sin \left( \pi -\dfrac{\pi }{4} \right)$ in the above expression. Then,
$\begin{aligned} & -2\sin \dfrac{3\pi }{4}\sin x \\\ & \Rightarrow -2\sin \left( \pi -\dfrac{\pi }{4} \right)\sin x \\\ \end{aligned}$
Now, we will use the formula from the equation (3) to write $\sin \left( \pi -\dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{4}$ in the above expression. Then,
$\begin{aligned} & -2\sin \left( \pi -\dfrac{\pi }{4} \right)\sin x \\\ & \Rightarrow -2\sin \dfrac{\pi }{4}\sin x \\\ \end{aligned}$
Now, we will use the formula from the equation (4) to write $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ in the above expression. Then,
$\begin{aligned} & -2\sin \dfrac{\pi }{4}\sin x \\\ & \Rightarrow -2\times \dfrac{1}{\sqrt{2}}\times \sin x \\\ & \Rightarrow -\sqrt{2}\sin x \\\ \end{aligned}$
Now, from the above result, we conclude that the value of the expression $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)$ will be equal to the value of the expression $-\sqrt{2}\sin x$ . Then,
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x$
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, $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x$ .
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 of $\cos \left( A+B \right)$ and $\cos \left( A-B \right)$ correctly. Moreover, we could have proved the result by applying the formula $\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ to write $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\sin \dfrac{3\pi }{4}\sin x$ directly.