Question

1. determine the exact value of each trigonometric ratio.

a) ( cos 75^circ )
b) ( \tan (-15^circ) )
c) ( cos \frac{11pi}{12} )
d) ( sin \frac{13pi}{12} )
e) ( \tan \frac{7pi}{12} )
f) ( \tan \frac{-5pi}{12} )

Explanation:

Step1: 拆分75°为特殊角

$\cos75^\circ=\cos(45^\circ+30^\circ)$

Step2: 用余弦和角公式展开

$\cos(A+B)=\cos A\cos B-\sin A\sin B$
$\cos75^\circ=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ$

Step3: 代入特殊角三角函数值

$\cos45^\circ=\frac{\sqrt{2}}{2},\cos30^\circ=\frac{\sqrt{3}}{2},\sin45^\circ=\frac{\sqrt{2}}{2},\sin30^\circ=\frac{1}{2}$
$\cos75^\circ=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$

Step4: 利用正切奇偶性化简

$\tan(-15^\circ)=-\tan15^\circ$，$15^\circ=45^\circ-30^\circ$

Step5: 用正切差角公式展开

$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$
$\tan15^\circ=\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ}$

Step6: 代入特殊角三角函数值

$\tan45^\circ=1,\tan30^\circ=\frac{\sqrt{3}}{3}$
$\tan15^\circ=\frac{1-\frac{\sqrt{3}}{3}}{1+1\times\frac{\sqrt{3}}{3}}=\frac{3-\sqrt{3}}{3+\sqrt{3}}=2-\sqrt{3}$，故$\tan(-15^\circ)=-(2-\sqrt{3})=\sqrt{3}-2$

Step7: 拆分$\frac{11\pi}{12}$为特殊角

$\cos\frac{11\pi}{12}=\cos(\pi-\frac{\pi}{12})=-\cos\frac{\pi}{12}$，$\frac{\pi}{12}=15^\circ=45^\circ-30^\circ$

Step8: 用余弦差角公式展开

$\cos\frac{\pi}{12}=\cos(45^\circ-30^\circ)=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ$

Step9: 代入特殊角三角函数值

$\cos\frac{\pi}{12}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$，故$\cos\frac{11\pi}{12}=-\frac{\sqrt{6}+\sqrt{2}}{4}$

Step10: 拆分$\frac{13\pi}{12}$为特殊角

$\sin\frac{13\pi}{12}=\sin(\pi+\frac{\pi}{12})=-\sin\frac{\pi}{12}$

Step11: 用正弦差角公式展开

$\sin\frac{\pi}{12}=\sin(45^\circ-30^\circ)=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ$

Step12: 代入特殊角三角函数值

$\sin\frac{\pi}{12}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$，故$\sin\frac{13\pi}{12}=-\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{2}-\sqrt{6}}{4}$

Step13: 拆分$\frac{7\pi}{12}$为特殊角

$\tan\frac{7\pi}{12}=\tan(\frac{\pi}{3}+\frac{\pi}{4})$

Step14: 用正切和角公式展开

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

Step15: 代入特殊角三角函数值

$\tan\frac{\pi}{3}=\sqrt{3},\tan\frac{\pi}{4}=1$
$\tan\frac{7\pi}{12}=\frac{\sqrt{3}+1}{1-\sqrt{3}\times1}=\frac{1+\sqrt{3}}{1-\sqrt{3}}=-2-\sqrt{3}$

Step16: 利用正切奇偶性化简

$\tan(-\frac{5\pi}{12})=-\tan\frac{5\pi}{12}$，$\frac{5\pi}{12}=\frac{\pi}{6}+\frac{\pi}{4}$

Step17: 用正切和角公式展开

$\tan\frac{5\pi}{12}=\frac{\tan\frac{\pi}{6}+\tan\frac{\pi}{4}}{1-\tan\frac{\pi}{6}\tan\frac{\pi}{4}}$

Step18: 代入特殊角三角函数值

$\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3},\tan\frac{\pi}{4}=1$
$\tan\frac{5\pi}{12}=\frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}\times1}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=2+\sqrt{3}$，故$\tan(-\frac{5\pi}{12})=-(2+\sqrt{3})=-2-\sqrt{3}$

Answer:

a) $\boldsymbol{\frac{\sqrt{6}-\sqrt{2}}{4}}$
b) $\boldsymbol{\sqrt{3}-2}$
c) $\boldsymbol{-\frac{\sqrt{6}+\sqrt{2}}{4}}$
d) $\boldsymbol{\frac{\sqrt{2}-\sqrt{6}}{4}}$
e) $\boldsymbol{-2-\sqrt{3}}$
f) $\boldsymbol{-2-\sqrt{3}}$