question 5\nfind the exact value of cos(13π/12).\nsubmit question

Explanation:

Step1: Rewrite the angle

We can write $\frac{13\pi}{12}=\frac{3\pi}{12}+\frac{10\pi}{12}=\frac{\pi}{4}+\frac{5\pi}{6}$.

Step2: Use the cosine - sum formula

The cosine - sum formula is $\cos(A + B)=\cos A\cos B-\sin A\sin B$. Here $A=\frac{\pi}{4}$ and $B = \frac{5\pi}{6}$. We know that $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$, $\sin\frac{5\pi}{6}=\frac{1}{2}$.

Step3: Substitute the values

$\cos(\frac{\pi}{4}+\frac{5\pi}{6})=\cos\frac{\pi}{4}\cos\frac{5\pi}{6}-\sin\frac{\pi}{4}\sin\frac{5\pi}{6}$. $=\frac{\sqrt{2}}{2}\times(-\frac{\sqrt{3}}{2})-\frac{\sqrt{2}}{2}\times\frac{1}{2}$. $=-\frac{\sqrt{6}+\sqrt{2}}{4}$.

Answer:

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