Question

Prove that $cos(\frac{3{\pi}}{4}+x)-cos(\frac{3{\pi}}{4}-x)=√2sin\,x$

Answer 1

It is known that $cosA-cosB=-2(\frac{A+B}{2}).sin(\frac{A-B}{2}).$

$∴L.H.S. = cos(\frac{3{\pi}}{4}-x) -cos(\frac{3{\pi}}{4}-x)$

$=-2\,sin\\{\frac{\,{(\frac{3{\pi}}{4}+x})+{(\frac{3{\pi}}{4}-x)}}{2}\\}.sin\\{\frac{\,{(\frac{3{\pi}}{4}+x})+{(\frac{3{\pi}}{4}-x)}}{2}\\}$

$=-2\,sin(\frac{3{\pi}}{4})sin\,x$

$=-2\,sin({\pi}-\frac{\pi}{4})sin\,x$

$=-2\,sin\frac{\pi}{4}sin\,x$

$=-2×\frac{1}{\sqrt2}×sin\,x$

$=-{\sqrt2}\,sin\,x$

$=R.H.S$