find the exact values of the sine, cosine, and tangent of the angle. 13π/12 sin(13π/12) = cos(13π/12) =…

Explanation:

Step1: Rewrite $\frac{13\pi}{12}$

Rewrite $\frac{13\pi}{12}=\pi+\frac{\pi}{12}$. Then use the sum - angle formulas $\sin(A + B)=\sin A\cos B+\cos A\sin B$, $\cos(A + B)=\cos A\cos B-\sin A\sin B$ and $\tan(A + B)=\frac{\tan A+\tan B}{1 - \tan A\tan B}$. Also, $\sin(\pi+\alpha)=-\sin\alpha$, $\cos(\pi+\alpha)=-\cos\alpha$, $\tan(\pi+\alpha)=\tan\alpha$. First, find $\sin\frac{\pi}{12}=\sin(\frac{\pi}{3}-\frac{\pi}{4})=\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}-\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$, $\cos\frac{\pi}{12}=\cos(\frac{\pi}{3}-\frac{\pi}{4})=\cos\frac{\pi}{3}\cos\frac{\pi}{4}+\sin\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{1}{2}\times\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$, $\tan\frac{\pi}{12}=\frac{\sin\frac{\pi}{12}}{\cos\frac{\pi}{12}} = 2-\sqrt{3}$. Since $\sin\frac{13\pi}{12}=\sin(\pi+\frac{\pi}{12})=-\sin\frac{\pi}{12}=-\frac{\sqrt{6}-\sqrt{2}}{4}$, $\cos\frac{13\pi}{12}=\cos(\pi+\frac{\pi}{12})=-\cos\frac{\pi}{12}=-\frac{\sqrt{6}+\sqrt{2}}{4}$, $\tan\frac{13\pi}{12}=\tan(\pi+\frac{\pi}{12})=\tan\frac{\pi}{12}=2 - \sqrt{3}$.

Step2: Rewrite $345^{\circ}$

Rewrite $345^{\circ}=360^{\circ}-15^{\circ}$. Then $\sin345^{\circ}=\sin(360^{\circ}-15^{\circ})=-\sin15^{\circ}$, $\cos345^{\circ}=\cos(360^{\circ}-15^{\circ})=\cos15^{\circ}$, $\tan345^{\circ}=\tan(360^{\circ}-15^{\circ})=-\tan15^{\circ}$. And $\sin15^{\circ}=\sin(45^{\circ}-30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$, $\cos15^{\circ}=\cos(45^{\circ}-30^{\circ})=\cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$, $\tan15^{\circ}=2 - \sqrt{3}$. So $\sin345^{\circ}=-\frac{\sqrt{6}-\sqrt{2}}{4}$, $\cos345^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$, $\tan345^{\circ}=-(2 - \sqrt{3})=\sqrt{3}-2$.

Answer:

$\sin(\frac{13\pi}{12})=-\frac{\sqrt{6}-\sqrt{2}}{4}$ $\cos(\frac{13\pi}{12})=-\frac{\sqrt{6}+\sqrt{2}}{4}$ $\tan(\frac{13\pi}{12})=2 - \sqrt{3}$ $\sin(345^{\circ})=-\frac{\sqrt{6}-\sqrt{2}}{4}$ $\cos(345^{\circ})=\frac{\sqrt{6}+\sqrt{2}}{4}$ $\tan(345^{\circ})=\sqrt{3}-2$