Question

Prove that $\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)=\sin (x+y)$

Answer 1

$\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)-\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)$

$=\frac{1}{2}\left[2\cos (\frac{\pi }{4}-x)\cos (\frac{\pi }{4}-y)\right]+\frac{1}{2}[-2\sin (\frac{\pi }{4}-x)\sin (\frac{\pi }{4}-y)]$

$=\frac{1}{2}[\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}+\cos \{(\frac{\pi }{4}-x)-(\frac{\pi }{4}-y)\}]$

$+\frac{1}{2}[\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}-\cos \{\frac{\pi }{4}-x\}-(\frac{\pi }{4}-y)\}$

$\left[\begin{array}{c}\because 2\cos B=\cos (A+B)+\cos (A-B)\\ -2\sin A\sin B=\cos (A+B)-\cos (A-B)\end{array}\right]$

$=2\times \frac{1}{2}\left[\cos \{(\frac{\pi }{4}-x)+(\frac{\pi }{4}-y)\}\right]$

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

$=\sin (x+y)$

$=R.H.S$