find the exact values of the sine, cosine, and tangent of the angle.\n17π/12 = 9π/4 - 5π/6\nsin(17π/12) =…

Explanation:

Step1: Recall angle - difference formulas

$\sin(A - B)=\sin A\cos B-\cos A\sin B$, $\cos(A - B)=\cos A\cos B+\sin A\sin B$, $\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$. Let $A=\frac{9\pi}{4}$ and $B = \frac{5\pi}{6}$. We know that $\sin\frac{9\pi}{4}=\sin(2\pi+\frac{\pi}{4})=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\cos\frac{9\pi}{4}=\cos(2\pi+\frac{\pi}{4})=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\tan\frac{9\pi}{4}=\tan(2\pi+\frac{\pi}{4})=\tan\frac{\pi}{4}=1$. Also, $\sin\frac{5\pi}{6}=\frac{1}{2}$, $\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$, $\tan\frac{5\pi}{6}=-\frac{\sqrt{3}}{3}$.

Step2: Calculate $\cos(\frac{17\pi}{12})$

$\cos(\frac{17\pi}{12})=\cos(\frac{9\pi}{4}-\frac{5\pi}{6})=\cos\frac{9\pi}{4}\cos\frac{5\pi}{6}+\sin\frac{9\pi}{4}\sin\frac{5\pi}{6}$ $=\frac{\sqrt{2}}{2}\times(-\frac{\sqrt{3}}{2})+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{2}-\sqrt{6}}{4}$.

Step3: Calculate $\tan(\frac{17\pi}{12})$

$\tan(\frac{17\pi}{12})=\tan(\frac{9\pi}{4}-\frac{5\pi}{6})=\frac{\tan\frac{9\pi}{4}-\tan\frac{5\pi}{6}}{1 + \tan\frac{9\pi}{4}\tan\frac{5\pi}{6}}$ $=\frac{1-(-\frac{\sqrt{3}}{3})}{1+1\times(-\frac{\sqrt{3}}{3})}=\frac{3 + \sqrt{3}}{3-\sqrt{3}}$. Rationalize the denominator: $\frac{(3 + \sqrt{3})(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}=\frac{9 + 6\sqrt{3}+3}{9 - 3}=\frac{12 + 6\sqrt{3}}{6}=2+\sqrt{3}$.

Answer:

$\cos(\frac{17\pi}{12})=\frac{\sqrt{2}-\sqrt{6}}{4}$, $\tan(\frac{17\pi}{12})=2+\sqrt{3}$