12. -/6 points details my notes larpcalclimaga8 5.4.018. 0/65 find the exact values of the sine, cosine, and…

Explanation:

Step1: Use angle - difference formula for sine

We know that $\sin(A - B)=\sin A\cos B-\cos A\sin B$. Here $A = \frac{2\pi}{3}$ and $B=\frac{7\pi}{4}$. $\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}$, $\cos\frac{2\pi}{3}=-\frac{1}{2}$, $\sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}$, $\cos\frac{7\pi}{4}=\frac{\sqrt{2}}{2}$. $\sin\left(-\frac{13\pi}{12}\right)=\sin\left(\frac{2\pi}{3}-\frac{7\pi}{4}\right)=\sin\frac{2\pi}{3}\cos\frac{7\pi}{4}-\cos\frac{2\pi}{3}\sin\frac{7\pi}{4}$ $=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}-\left(-\frac{1}{2}\right)\times\left(-\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{6}-\sqrt{2}}{4}$

Step2: Use angle - difference formula for cosine

We know that $\cos(A - B)=\cos A\cos B+\sin A\sin B$. $\cos\left(-\frac{13\pi}{12}\right)=\cos\left(\frac{2\pi}{3}-\frac{7\pi}{4}\right)=\cos\frac{2\pi}{3}\cos\frac{7\pi}{4}+\sin\frac{2\pi}{3}\sin\frac{7\pi}{4}$ $=-\frac{1}{2}\times\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\times\left(-\frac{\sqrt{2}}{2}\right)=-\frac{\sqrt{2}+\sqrt{6}}{4}$

Step3: Use the definition of tangent

$\tan\theta=\frac{\sin\theta}{\cos\theta}$. $\tan\left(-\frac{13\pi}{12}\right)=\frac{\sin\left(-\frac{13\pi}{12}\right)}{\cos\left(-\frac{13\pi}{12}\right)}=\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{-\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$ Rationalize the denominator: Multiply numerator and denominator by $\sqrt{6}-\sqrt{2}$ $\tan\left(-\frac{13\pi}{12}\right)=2 + \sqrt{3}$

Answer:

$\sin\left(-\frac{13\pi}{12}\right)=\frac{\sqrt{6}-\sqrt{2}}{4}$, $\cos\left(-\frac{13\pi}{12}\right)=-\frac{\sqrt{6}+\sqrt{2}}{4}$, $\tan\left(-\frac{13\pi}{12}\right)=2+\sqrt{3}$