Question

If $A={60}^0$ and $B={30}^0$, 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 $B={30}^{\circ }$

= ${(\sin 60\cos 30+\cos 60\sin 30)}^2+{(\cos 60\cos 30-\sin 60\sin 30)}^2$

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

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

= $1^2$

= $1$