Question

Find the exact value of the expression:

$\sin \left(\frac{\mathrm{\pi }}{12}\right)$

Short answer

$\frac{1}{4}(\sqrt{6}-\sqrt{2})$.

Step-by-step solution

Step 1. Given information 

We have been given an expression $\sin \left(\frac{\mathrm{\pi }}{12}\right)$.

We have to find its exact value.

Step 2. Simplify the expression

We can also write the expression as:

$$\begin{gathered} \sin \left(\frac{\mathrm{\pi }}{12}\right)=\sin \left(\left(\frac{3\mathrm{\pi }}{12}\right)-\left(\frac{2\mathrm{\pi }}{12}\right)\right) \\ =\sin \left(\left(\frac{\mathrm{\pi }}{4}\right)-\left(\frac{\mathrm{\pi }}{6}\right)\right) \end{gathered}$$

Step 3.  Apply the compound angle formula

$\sin \left(\left(\frac{\mathrm{\pi }}{4}\right)-\left(\frac{\mathrm{\pi }}{6}\right)\right)$has the form $\sin \left(\alpha -\beta \right)$

Therefore, apply

$\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$

$\begin{array}{r}\sin \left(\frac{\pi }{4}\right)\cos \left(\frac{\pi }{6}\right)-\cos \left(\frac{\pi }{4}\right)\sin \left(\frac{\pi }{6}\right)=\left(\left(\frac{\sqrt{2}}{2}\right)\cdot \left(\frac{\sqrt{3}}{2}\right)\right)-\left(\left(\frac{\sqrt{2}}{2}\right)\cdot \left(\frac{1}{2}\right)\right)\\ =\left(\frac{\sqrt{6}}{4}\right)-\left(\frac{\sqrt{2}}{4}\right)\\ =\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)\\ =\frac{1}{4}(\sqrt{6}-\sqrt{2})\end{array}$