find the exact values of the sine, cosine, and tangent of the angle. \n\frac{11pi}{12}=\frac{3pi}{4}+\frac{pi…

Explanation:

Step1: Recall sum - of - angles formula for sine

$\sin(A + B)=\sin A\cos B+\cos A\sin B$. Here $A=\frac{3\pi}{4}$ and $B = \frac{\pi}{6}$. We know that $\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}$, $\cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}$, $\sin\frac{\pi}{6}=\frac{1}{2}$, $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$. $\sin(\frac{11\pi}{12})=\sin(\frac{3\pi}{4}+\frac{\pi}{6})=\sin\frac{3\pi}{4}\cos\frac{\pi}{6}+\cos\frac{3\pi}{4}\sin\frac{\pi}{6}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+(-\frac{\sqrt{2}}{2})\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$

Step2: Recall sum - of - angles formula for cosine

$\cos(A + B)=\cos A\cos B-\sin A\sin B$. $\cos(\frac{11\pi}{12})=\cos(\frac{3\pi}{4}+\frac{\pi}{6})=\cos\frac{3\pi}{4}\cos\frac{\pi}{6}-\sin\frac{3\pi}{4}\sin\frac{\pi}{6}=(-\frac{\sqrt{2}}{2})\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=-\frac{\sqrt{6}+\sqrt{2}}{4}$

Step3: Recall the formula for tangent

$\tan\theta=\frac{\sin\theta}{\cos\theta}$. $\tan(\frac{11\pi}{12})=\frac{\sin(\frac{11\pi}{12})}{\cos(\frac{11\pi}{12})}=\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{-\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}=\frac{(\sqrt{6}-\sqrt{2})^2}{-((\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2}))}=\frac{6 - 2\sqrt{12}+2}{-(6 - 2)}=\frac{8 - 4\sqrt{3}}{-4}= \sqrt{3}- 2$

Answer:

$\sin(\frac{11\pi}{12})=\frac{\sqrt{6}-\sqrt{2}}{4}$, $\cos(\frac{11\pi}{12})=-\frac{\sqrt{6}+\sqrt{2}}{4}$, $\tan(\frac{11\pi}{12})=\sqrt{3}-2$