Question

If $A=$ ${60}^{\circ }$ and $B=$ ${30}^{\circ }$, find the value of $(\sin A\cos B+\cos A\sin B{)}^2+(\cos A\cos B-\sin A\sin B{)}^2$.

Answer 1

${(\sin A\cos B+\cos A\sin B)}^2+{(\cos A\cos B-\sin A\sin B)}^2$
When, $A={60}^{\circ }$ and $B={30}^{\circ }$
$={(\sin {60}^0\cos {30}^0+\cos {60}^0\sin {30}^0)}^2+{(\cos {60}^0\cos {30}^0-\sin {60}^0\sin {30}^0)}^2$

$={(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2})}^2+{(\frac{1}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times \frac{1}{2})}^2$

$={(\frac{3}{4}+\frac{1}{4})}^2$
$=1^2=1$