Question

Find the exact values of sin pi/12

Answer 1

Answer:

$\sin (\frac{\pi}{12})=\frac{\sqrt{6}-\sqrt{2}}{4}$

Step-by-step explanation:

We want to find the exact value of

$\sin \frac{\pi}{12}$

We rewrite the given expression to obtain:

$\sin \frac{\pi}{12}=\sin (\frac{\pi}{3}-\frac{\pi}{4})$

We use the identity:

$\sin (A-B)=\sin A \cos B-\sin B \cos A$

$\sin ( \frac{\pi}{3}- \frac{\pi}{4})=\sin \frac{\pi}{3} \cos \frac{\pi}{4}-\sin \frac{\pi}{4} \cos \frac{\pi}{3}$

$\sin ( \frac{\pi}{3}- \frac{\pi}{4})=\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2} \times \frac{1}{2}$

$\sin ( \frac{\pi}{3}- \frac{\pi}{4})=\frac{\sqrt{6}-\sqrt{2}}{4}$

Answer 2

$0$ and $\sin x$ is monotonic on $0$, so we know

$\sin0$

The important thing here is that $\sin\dfrac\pi{12}$ must be positive.

Recall the half-angle identity:

$\sin^2\dfrac\pi{12}=\dfrac{1-\cos\frac\pi6}2\implies\sin\dfrac\pi{12}=\sqrt{\dfrac{1-\cos\frac\pi6}2}$

(as opposed to the negative square root)

So we have

$\sin\dfrac\pi{12}=\sqrt{\dfrac{1-\frac{\sqrt3}2}2}=\sqrt{\dfrac12-\dfrac{\sqrt3}4}=\dfrac{\sqrt3-1}{2\sqrt2}$