Question

Derive:$\cos x+\cos y=2\cos \frac{\left(x+y\right)}{2}\cos \frac{\left(x-y\right)}{2}$

Answer 1

$R.H.S.$

$2\cos \frac{(x+y)}{2}\cos \frac{(x-y)}{2}$

$=2\cos (\frac{x}{2}+\frac{y}{2})\cos (\frac{x}{2}-\frac{y}{2})$

$=2[\cos \frac{x}{2}\cos \frac{y}{2}-\sin \frac{x}{2}\sin \frac{y}{2}][\cos \frac{x}{2}\cos \frac{y}{2}+\sin \frac{x}{2}\sin \frac{y}{2}]$

$=2[{\left(\cos \frac{x}{2}\cos \frac{y}{2}\right)}^2-{\left(\sin \frac{x}{2}\sin \frac{y}{2}\right)}^2]$

$=2[{\cos }^2\frac{x}{2}{\cos }^2\frac{y}{2}-{\sin }^2\frac{x}{2}{\sin }^2\frac{y}{2}]$

$=2[{\cos }^2\frac{x}{2}(1-{\sin }^2\frac{y}{2})-(1-{\cos }^2\frac{x}{2}){\sin }^2\frac{y}{2}]$

$=2[{\cos }^2\frac{x}{2}-{\cos }^2\frac{x}{2}{\sin }^2\frac{y}{2}-{\sin }^2\frac{y}{2}+{\sin }^2\frac{y}{2}{\cos }^2\frac{x}{2}]$

$=2[{\cos }^2\frac{x}{2}-{\sin }^2\frac{y}{2}]$

$=2[\frac{1+\cos x}{2}-\frac{1-\cos y}{2}]$

$=\frac{2}{2}[1+\cos x-1+\cos y]$

$=\cos x-\cos y$

$L.H.S.$

Hence proved.