Prove that: \[ \cos\left(\frac{3\pi}{4}+x\right) – \cos\left(\frac{3\pi}{4}-x\right) = -\sqrt{2}\sin x \]
Solution
Using the identity:
\[
\cos A – \cos B
=
-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
\]
Taking
\[
A=\frac{3\pi}{4}+x,
\qquad
B=\frac{3\pi}{4}-x
\]
Then,
\[
\cos\left(\frac{3\pi}{4}+x\right)
–
\cos\left(\frac{3\pi}{4}-x\right)
\]
\[
=
-2\sin\frac{\left(\frac{3\pi}{4}+x\right)+\left(\frac{3\pi}{4}-x\right)}{2}
\sin\frac{\left(\frac{3\pi}{4}+x\right)-\left(\frac{3\pi}{4}-x\right)}{2}
\]
\[
=
-2\sin\frac{6\pi/4}{2}\sin\frac{2x}{2}
\]
\[
=
-2\sin\frac{3\pi}{4}\sin x
\]
\[
=
-2\times\frac{\sqrt{2}}{2}\sin x
\]
\[
=
-\sqrt{2}\sin x
\]
Hence,
\[
\boxed{
\cos\left(\frac{3\pi}{4}+x\right)
–
\cos\left(\frac{3\pi}{4}-x\right)
=
-\sqrt{2}\sin x
}
\]