12 mark for review the function f is given by f(θ)=sinθ. using the fact that π/12 = π/4+(-π/6), what is the…

Explanation:

Step1: Use sine - addition formula

The sine - addition formula is $\sin(A + B)=\sin A\cos B+\cos A\sin B$. Here $A=\frac{\pi}{4}$ and $B =-\frac{\pi}{6}$, and we want to find $\sin(\frac{\pi}{12})=\sin(\frac{\pi}{4}+(-\frac{\pi}{6}))$.

Step2: Recall trigonometric values

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

Step3: Substitute values into formula

$\sin(\frac{\pi}{4}+(-\frac{\pi}{6}))=\sin\frac{\pi}{4}\cos(-\frac{\pi}{6})+\cos\frac{\pi}{4}\sin(-\frac{\pi}{6})$. Substitute the values: $\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times(-\frac{1}{2})$.

Step4: Simplify the expression

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

Answer:

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