Question

Find the value of$sin{75}^0$

Answer 1

$\sin {75}^{\circ }=\sin ({30}^{\circ }+{45}^{\circ })\sin (A+B)=\sin A\cos B+\cos A\sin B\sin {75}^{\circ }=\sin ({30}^{\circ }+{45}^{\circ })=\sin {30}^{\circ }\cos {45}^{\circ }+\cos {30}^{\circ }\sin {45}^{\circ }=\frac{1}{2}\times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$