QPaperGen

We use the identity:


\[
\cos A - \cos B = -2 \sin\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right)
\]



Let:


\[
A = \frac{3\pi}{4} + x, \quad B = \frac{3\pi}{4} - x
\]



Then:
\begin{align*}
\text{LHS} &= \cos\left( \frac{3\pi}{4} + x \right) - \cos\left( \frac{3\pi}{4} - x \right) \\
&= -2 \sin\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right) \\
&= -2 \sin\left( \frac{ \left( \frac{3\pi}{4} + x \right) + \left( \frac{3\pi}{4} - x \right) }{2} \right) \sin\left( \frac{ \left( \frac{3\pi}{4} + x \right) - \left( \frac{3\pi}{4} - x \right) }{2} \right) \\
&= -2 \sin\left( \frac{3\pi}{4} \right) \sin x
\end{align*}

Now:


\[
\sin\left( \frac{3\pi}{4} \right) = \sin\left( \pi - \frac{\pi}{4} \right) = \sin\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}
\]



So:
\begin{align*}
\text{LHS} &= -2 \cdot \frac{1}{\sqrt{2}} \cdot \sin x \\
&= -\sqrt{2} \sin x = \text{RHS} \qquad \textbf{Hence proved.}
\end{align*}