Question

Find the value of $\sin {75}^{\circ }$.

Answer 1

Find the value of given trigonometric ratio.

Given trigonometric ratio: $\sin {75}^{\circ }$

$\sin {75}^{\circ }$can be expressed as, $\sin {75}^{\circ }=\sin ({45}^{\circ }+{30}^{\circ })$

We know that $$\sin (A+B)=\sin A\cos B+\cos A\sin B$$

So by applying the above formula we get,

$\sin {75}^{\circ }=\sin {45}^{\circ }\cos {30}^{\circ }+\cos {45}^{\circ }\sin {30}^{\circ }$

As, $\sin {45}^{\circ }=\frac{1}{\sqrt{2}},\cos {30}^{\circ }=\frac{\sqrt{3}}{2}=,\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$ and $\sin {30}^{\circ }=\frac{1}{2}$.

By substituting the values we get,

$$\begin{gathered} \sin {75}^{\circ }=\left(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\times \frac{1}{2}\right) \\ =\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \\ =\frac{\sqrt{3}+1}{2\sqrt{2}} \end{gathered}$$

Hence, the value of $\sin {75}^{\circ }$is $\frac{\sqrt{3}+1}{2\sqrt{2}}$.