Question

Find the value of $\sin {15}^{\circ }$.

Answer 1

Find the value of the trigonometric function using identities :

Given trigonometric function: $\sin {15}^{\circ }$.

$\sin {15}^{\circ }$can be expressed as, $\sin {15}^{\circ }=\sin ({45}^{\circ }-{30}^{\circ })$

Using the identity $\sin (A-B)=\sin A\cos B-\cos A\sin B$ we can write,

$\sin ({45}^{\circ }-{30}^{\circ })=\sin {45}^{\circ }\times \cos {30}^{\circ }-\cos {45}^{\circ }\times \sin {30}^{\circ }$

We know that $\sin {45}^{\circ }=\frac{1}{\sqrt{2}},\sin {30}^{\circ }=\frac{1}{2},\cos {45}^{\circ }=\frac{1}{\sqrt{2}},$ and $\cos {30}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$

By substituting the above values we get,

$$\begin{gathered} \sin ({45}^{\circ }-{30}^{\circ })=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \\ =\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}} \\ =\frac{\sqrt{3}-1}{2\sqrt{2}} \\ \therefore \sin {15}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}} \end{gathered}$$
Hence, the value of $\sin {15}^{\circ }$is$\frac{\sqrt{3}-1}{2\sqrt{2}}$.