Question

State true or false.

$(\cos x+\mathrm{cosy}{)}^2+(\sin x-\sin y{)}^2=4{\cos }^2\left(\frac{x+y}{2}\right)$

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

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

$=4{\cos }^2\left(\frac{x+y}{2}\right)\times 1$

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

$=R.H.S$

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