Question

Prove that: $(\cos A+\cos B{)}^2+(\sin A-\sin B{)}^2=4{\cos }^2\frac{A+B}{2}$.

Answer 1

$L.H.S.$

$={\left(cos\right(A)+cos(B\left)\right)}^2+{\left(sin\right(A)-sin(B\left)\right)}^2$

$=co{s}^2\left(A\right)+co{s}^2\left(B\right)+2cos\left(A\right)cos\left(B\right)+si{n}^2\left(A\right)+si{n}^2\left(B\right)-2sin\left(A\right)sin\left(B\right)$

$=2+2\left(cos\right(A\left)cos\right(B)-sin(A\left)sin\right(B\left)\right)$

$=2+2cos(A+B)$

$=2+2(2co{s}^2\left(\frac{A+B}{2}\right)-1)$

$=4co{s}^2\left(\frac{A+B}{2}\right)$

$=R.H.S.$