Question

Prove that:

\[\cos\left( \frac{\pi}{4} + x \right) + \cos\left( \frac{\pi}{4} - x \right) = \sqrt{2} \cos x\]

 

Answer 1

Consider LHS:
\[\cos \left( \frac{\pi}{4} + x \right) + \cos\left( \frac{\pi}{4} - x \right)\]
\[ = 2\cos \left\{ \frac{\left( \frac{\pi}{4} + x \right) + \left( \frac{\pi}{4} - x \right)}{2} \right\}\cos \left\{ \frac{\left( \frac{\pi}{4} + x \right) - \left( \frac{\pi}{4} + x \right)}{2} \right\} \left\{ \because \cos A + \cos B = 2\cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right\}\]
\[= 2\cos \left\{ \frac{\frac{\pi}{4} + x + \frac{\pi}{4} - x}{2} \right\}\cos \left\{ \frac{\frac{\pi}{4} + x - \frac{\pi}{4} + x}{2} \right\}\]
\[ = 2\cos$\left( \frac{\pi}{4} \right)$ \cos x\]
\[ = 2 \times \frac{1}{\sqrt{2}} \times \cos x\]
\[ = \sqrt{2}\cos x\]
 = RHS
Hence, LHS = RHS