求其他的三角函数值
个人记录
特殊三角函数值
| $\alpha = 30^{\circ}$ | $\alpha = 45^{\circ}$ | $\alpha = 60^{\circ}$ |
|---|---|---|
| $\sin \alpha=\displaystyle \frac{1}{2}$ | $\sin \alpha = \displaystyle \frac{\sqrt{2}}{2}$ | $\sin \alpha = \displaystyle \frac{\sqrt{3}}{2}$ |
| $\cos \alpha=\displaystyle \frac{\sqrt{3}}{2}$ | $\cos \alpha = \displaystyle \frac{\sqrt{2}}{2}$ | $\cos \alpha=\displaystyle \frac{1}{2}$ |
| $\tan \alpha=\sqrt{3}$ | $\tan \alpha =1$ | $\tan \alpha=\displaystyle \frac{\sqrt{3}}{3}$ |
用到的公式
$\sin(\alpha + \beta)=\sin\alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta$
$\cos(\alpha+\beta)=\cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta$
$\tan(\alpha+\beta)=\displaystyle \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \cdot \tan \beta}$
$\sin(\alpha - \beta)=\sin\alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta$
$\cos(\alpha - \beta)=\cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta$
$\tan(\alpha - \beta)=\displaystyle \frac{\tan \alpha - \tan \beta}{1+\tan \alpha \cdot \tan \beta}$
开始狂K其他三角函数值
- 求 $15^\circ$ 三角函数值。
令 $\alpha=60^\circ,\beta=45^\circ$，那么可得 $\sin(\alpha - \beta)=\sin\alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta=\sin(15^\circ)=\displaystyle \frac{\sqrt{3}}{2} \cdot \displaystyle \frac{\sqrt{2}}{2}-\displaystyle \frac{1}{2} \cdot \displaystyle \frac{\sqrt{2}}{2}= \displaystyle \frac{\sqrt{6}}{4}-\displaystyle \frac{\sqrt{2}}{4}=\displaystyle \frac{\sqrt{6}-\sqrt{2}}{4}$
$\cos(\alpha - \beta)=\cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta=\displaystyle \frac{1}{2} \cdot \displaystyle \frac{\sqrt{2}}{2}+\displaystyle \frac{\sqrt{3}}{2} \cdot \displaystyle \frac{\sqrt{2}}{2}=\displaystyle \frac{\sqrt{2}}{4} + \displaystyle \frac{\sqrt{6}}{4}=\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}$
$\tan(\alpha - \beta)=\displaystyle \frac{\tan \alpha - \tan \beta}{1+\tan \alpha \cdot \tan \beta}=\displaystyle \frac{\displaystyle \frac{\sqrt{3}}{3}- 1}{1+\displaystyle \frac{\sqrt{3}}{3}}=\displaystyle \frac{(\displaystyle \frac{\sqrt{3}}{3}-1)^2}{(1+\displaystyle \frac{\sqrt{3}}{3})\cdot (\displaystyle \frac{\sqrt{3}}{3}-1)}=\displaystyle \frac{\displaystyle \frac{4}{3}-\displaystyle \frac{2\sqrt{3}}{3}}{\displaystyle \frac{2}{3}}=\displaystyle \frac{\displaystyle \frac{4-2\sqrt{3}}{3}}{\displaystyle \frac{2}{3}}=\displaystyle \frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$ - 求 $75^{\circ}$ 的三角函数值。
由 $\sin \alpha = \cos(90^{\circ}-\alpha)$ 得 $\sin 75^{\circ} = \cos 15^{\circ} = \displaystyle \frac {\sqrt{2}+\sqrt{6}}{4}$
$\cos 75^{\circ} = \sin 15^{\circ}=\displaystyle \frac {\sqrt{6}-\sqrt{2}}{4}$
$\tan \alpha= \displaystyle \frac{\sin \alpha}{\cos \alpha}=\displaystyle \frac{\displaystyle \frac {\sqrt{2}+\sqrt{6}}{4}}{\displaystyle \frac {\sqrt{6}-\sqrt{2}}{4}}=\displaystyle \frac{\sqrt{2}+\sqrt{6}}{\sqrt{6}-\sqrt{2}}=\displaystyle \frac{(\sqrt{6}-\sqrt{2})^2}{(\sqrt{6}-\sqrt{2}) \cdot (\sqrt{6}+\sqrt{2})}=\displaystyle \frac {8-4\sqrt{3}}{4}=2-\sqrt{3}$