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