Question

Prove that ${(cosx+cosy)}^2+{(sinx-siny)}^2=4\left\{\frac{\cos }{}\right\}$

Answer 1

$L.H.S={\left(\cos x+\cos y\right)}^2+{\left(\sin x-\sin y\right)}^2$

We know that, ${\left(a+b\right)}^2=a^2+b^2+2ab$ and ${\left(a-b\right)}^2=a^2+b^2-2ab$.

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

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

$=1+1+2\cos \left(x+y\right)$……………………………………………………….(${\sin }^2\theta +{\cos }^2\theta =1\mathrm{and}\cos \left(\mathrm{a}+\mathrm{b}\right)=\mathrm{cosacosb}-\mathrm{sinasinb}$)

$=2+2\cos \left(x+y\right)$

$=2\left(1+\cos \left(x+y\right)\right)$

$=2\left[1+2{\cos }^2\left(\frac{x+y}{2}\right)-1\right]$……………($\cos 2A=2{\cos }^{2}A-1$)

$=2\left(2{\cos }^2\left(\frac{x+y}{2}\right)\right)$

$=4{\cos }^2\left(\frac{x+y}{2}\right)$.

$=R.H.S.$

Hence, it is proved that${(cosx+cosy)}^2+{(sinx-siny)}^2=4\left\{\frac{\cos }{}\right\}$