12. 2/6 points details my notes larpcalclimagab 5.4 find the exact values of the sine, cosine, and tangent…

Explanation:

Step1: Recall angle - difference formulas

We know that (\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}). Given (-\frac{13\pi}{12}=\frac{2\pi}{3}-\frac{7\pi}{4}), where (\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}), (\tan\frac{2\pi}{3}=-\sqrt{3}), (\tan\frac{7\pi}{4}=- 1).

Step2: Calculate (\sin(-\frac{13\pi}{12}))

(\sin(-\frac{13\pi}{12})=\sin(\frac{2\pi}{3}-\frac{7\pi}{4})=\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}-(-\frac{1}{2})\times(-\frac{\sqrt{2}}{2})=\frac{\sqrt{6}-\sqrt{2}}{4})

Step3: Calculate (\cos(-\frac{13\pi}{12}))

(\cos(-\frac{13\pi}{12})=\cos(\frac{2\pi}{3}-\frac{7\pi}{4})=\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(-\frac{\sqrt{2}}{2})=-\frac{\sqrt{6}+\sqrt{2}}{4})

Step4: Calculate (\tan(-\frac{13\pi}{12}))

(\tan(-\frac{13\pi}{12})=\tan(\frac{2\pi}{3}-\frac{7\pi}{4})=\frac{\tan\frac{2\pi}{3}-\tan\frac{7\pi}{4}}{1 + \tan\frac{2\pi}{3}\tan\frac{7\pi}{4}}=\frac{-\sqrt{3}-(-1)}{1+(-\sqrt{3})\times(-1)}=\frac{1 - \sqrt{3}}{1+\sqrt{3}}=\frac{(1 - \sqrt{3})(1 - \sqrt{3})}{(1+\sqrt{3})(1 - \sqrt{3})}=\frac{1-2\sqrt{3}+3}{1 - 3}=\frac{4 - 2\sqrt{3}}{-2}=\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})=\sqrt{3}-2)