find the exact value of sin $\frac{pi}{12}$ by using a sum or difference formula.

Explanation:

Step1: Rewrite $\frac{\pi}{12}$ as a difference

We know that $\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}$.

Step2: Apply the sine - difference formula

The sine - difference formula is $\sin(A - B)=\sin A\cos B-\cos A\sin B$. Here $A=\frac{\pi}{3}$ and $B = \frac{\pi}{4}$. So $\sin(\frac{\pi}{3}-\frac{\pi}{4})=\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}$.

Step3: Substitute the known values

We know that $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\cos\frac{\pi}{3}=\frac{1}{2}$, and $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$. Then $\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}-\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

Answer:

$\frac{\sqrt{6}-\sqrt{2}}{4}$