Question

How do you find the exact value of $\sin {105}^0$.

Answer 1

Solve for the value of given trigonometric ratio.

Given trigonometric ratio: $\sin {105}^0$

$\sin {105}^0$can be expressed as, $\sin {105}^{\circ }=\sin ({60}^{\circ }+{45}^{\circ })$

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

So by applying the above formula we get,

$\sin {105}^{\circ }=\sin ({60}^{\circ }+{45}^{\circ })=\sin {60}^{\circ }\cos {45}^{\circ }+\cos {60}^{\circ }\sin {45}^{\circ }$

As, $\sin {60}^{\circ }=\frac{\sqrt{3}}{2},\cos {45}^{\circ }=\frac{1}{\sqrt{2}},\cos {60}^{\circ }=\frac{1}{2}$ and $\sin {45}^{\circ }=\frac{1}{\sqrt{2}}$.

By substituting the values we get,

$$\begin{gathered} \sin {105}^{\circ }=\left(\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}\right)+\left(\frac{1}{2}\times \frac{1}{\sqrt{2}}\right) \\ =\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \\ =\frac{\sqrt{3}+1}{2\sqrt{2}} \end{gathered}$$

Hence, the exact value of $\sin {105}^{\circ }$is $\frac{\sqrt{3}+1}{2\sqrt{2}}$.