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烦人三角函数公式整理


弧、面积公式

\[\begin{aligned} l=\alpha \cdot r \qquad S=\frac12\alpha r^2=\frac12 l \, r \end{aligned} \]

升降幂公式

\[(\sin\alpha + \cos\alpha)^2 = 1 + 2\sin\alpha\ \cos\alpha \qquad (\sin\alpha - \cos\alpha)^2 = 1 - 2\sin\alpha\ \cos\alpha \]

诱导公式

\[\begin{aligned} \sin (2k \pi + \alpha) &= \sin \alpha & \cos (2k \pi + \alpha) &= \cos \alpha & \tan (2k \pi + \alpha) &= \tan \alpha\\ \end{aligned} \]

\[\begin{aligned} \sin ( \pi - \alpha) &= \sin \alpha & \cos ( \pi - \alpha) &= -\cos \alpha & \tan ( \pi - \alpha) &= -\tan \alpha\\ \sin ( - \alpha) &= -\sin \alpha & \cos ( - \alpha) &= \cos \alpha & \tan ( - \alpha) &= -\tan \alpha\\ \sin ( \pi + \alpha) &= -\sin \alpha & \cos ( \pi + \alpha) &= -\cos \alpha & \tan ( \pi + \alpha) &= \tan \alpha\\ \end{aligned} \]

\[\begin{aligned} \sin ( \frac\pi2 - \alpha) &= \cos \alpha & \sin ( \frac\pi2 + \alpha) &= \cos \alpha &\\ \cos ( \frac\pi2 - \alpha) &= \sin \alpha & \cos ( \frac\pi2 + \alpha) &= -\sin \alpha &\\ \tan ( \frac\pi2 - \alpha) &= \cot \alpha & \tan ( \frac\pi2 + \alpha) &= -\cot \alpha &\\ \cot ( \frac\pi2 - \alpha) &= \tan \alpha & \cot ( \frac\pi2 + \alpha) &= -\tan \alpha &\\ \end{aligned} \]

\[\begin{aligned} \sin ( \frac{3\pi}2 - \alpha) &= -\cos \alpha & \sin ( \frac{3\pi}2 + \alpha) &= -\cos \alpha &\\ \cos ( \frac{3\pi}2 - \alpha) &= -\sin \alpha & \cos ( \frac{3\pi}2 + \alpha) &= \sin \alpha &\\ \tan ( \frac{3\pi}2 - \alpha) &= -\cot \alpha & \tan ( \frac{3\pi}2 + \alpha) &= -\cot \alpha &\\ \cot ( \frac{3\pi}2 - \alpha) &= \tan \alpha & \cot ( \frac{3\pi}2 + \alpha) &= -\tan \alpha &\\ \end{aligned} \]

两角和差公式

\[\begin{aligned} \sin (\alpha + \beta) &= \sin\alpha\ \cos\beta + \cos\alpha\ \sin\beta & \sin (\alpha - \beta) &= \sin\alpha\ \cos\beta - \cos\alpha\ \sin\beta \\ \cos (\alpha + \beta) &= \cos\alpha\ \cos\beta - \sin\alpha\ \sin\beta & \cos (\alpha - \beta) &= \cos\alpha\ \cos\beta + \sin\alpha\ \sin\beta \\ \tan (\alpha + \beta) &= \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\ \tan\beta} & \tan (\alpha - \beta) &= \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\ \tan\beta} \\ \end{aligned} \]

倍角公式

\[\begin{aligned} \sin 2\alpha &= 2\sin\alpha\ \cos\alpha \\ \cos 2\alpha &= \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha-1 = 1-2\sin^2\alpha \\ \tan 2\alpha &= \frac{2\tan\alpha}{1 - \tan^2\alpha} \\ \end{aligned} \]

半角公式

\[\begin{aligned} \sin \frac\alpha2 &= \pm \sqrt{\frac{1 - \cos\alpha}{2}} \\ \cos \frac\alpha2 &= \pm \sqrt{\frac{1 + \cos\alpha}{2}} \\ \tan \frac\alpha2 &= \pm \sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}} \\ \tan \frac\alpha2 &= \frac{\sin\alpha}{1 + \cos\alpha}= \frac{1 - \cos\alpha}{\sin\alpha} \\ \end{aligned} \]

积化和差

\[\begin{aligned} \sin\alpha\ \cos\beta&=\frac12\left[\sin(\alpha+\beta)+\sin(\alpha-\beta)\right] & \cos\alpha\ \sin\beta&=\frac12\left[\sin(\alpha+\beta)-\sin(\alpha-\beta)\right] \\ \cos\alpha\ \cos\beta&=\frac12\left[\cos(\alpha+\beta)+\cos(\alpha-\beta)\right] & \sin\alpha\ \sin\beta&=-\frac12\left[\cos(\alpha+\beta)-\cos(\alpha-\beta)\right] \\ \end{aligned} \]

和差化积

\[\begin{aligned} \sin\alpha + \sin\beta&=2\sin\frac{\alpha+\beta}2\ \sin\frac{\alpha-\beta}2 & \sin\alpha - \sin\beta&=2\sin\frac{\alpha+\beta}2\ \sin\frac{\alpha-\beta}2 \\ \cos\alpha + \cos\beta&=2\cos\frac{\alpha+\beta}2\ \cos\frac{\alpha-\beta}2 & \cos\alpha - \cos\beta&=-2\cos\frac{\alpha+\beta}2\ \cos\frac{\alpha-\beta}2 \\ \end{aligned} \]

辅助角公式

\[a\sin\alpha + b\cos\alpha=\sqrt{a^2+b^2}\sin(\alpha+\varphi)\\ \left(\cos\varphi=\frac a{\sqrt{a^2+b^2}}, \ \sin\varphi=\frac b{\sqrt{a^2+b^2}}, \ \tan\varphi=\frac ba\right) \]

三倍角公式

\[\begin{aligned} \sin 3\alpha &= 3\sin\alpha - 4\sin^3\alpha = 4\sin\alpha\ \sin(\frac\pi3 - \alpha)\ \sin(\frac\pi3 + \alpha) \\ \cos 3\alpha &= 4\cos^3\alpha - 3\cos\alpha = 4\cos\alpha\ \cos(\frac\pi3 - \alpha)\ \cos(\frac\pi3 + \alpha) \\ \cos 3\alpha &= \frac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha} = 4\tan\alpha\ \tan(\frac\pi3 - \alpha)\ \tan(\frac\pi3 + \alpha) \\ \end{aligned} \]

正余切和差

\[\begin{aligned} \tan\alpha + \cot\alpha &= \frac1{\sin\alpha\ \cos\alpha} = \frac2{\sin 2\alpha}\\ \tan\alpha - \cot\alpha &= \frac{\sin^2\alpha\ - \cos^2\alpha}{\sin\alpha\ \cos\alpha} = \frac{-2\ \cos{2\alpha}}{\sin 2\alpha} = -2\ \cot{2\alpha}\\ \end{aligned} \]

万能公式

\[\begin{aligned} \sin\alpha &= \frac{2\tan\displaystyle\frac{\alpha}{2}}{1 + \tan^2\displaystyle\frac{\alpha}{2}} & \cos\alpha &= \frac{1 - \tan^2\displaystyle\frac{\alpha}{2}}{1 + \tan^2\displaystyle\frac{\alpha}{2}} & \tan\alpha &= \frac{2\tan\displaystyle\frac{\alpha}{2}}{1 - \tan^2\displaystyle\frac{\alpha}{2}} \\ \end{aligned} \]

降幂公式

\[\begin{aligned} \sin^2\alpha &= \frac{1 - \cos2\alpha}2 & \cos^2\alpha &= \frac{1 + \cos2\alpha}2 & \sin\alpha\ \cos\alpha &= \frac{\sin 2\alpha}2 \\ \end{aligned} \]

三角形恒等式

在三角形\(ABC\)中,\(A + B + C = 2\pi\)\(A、B、C\not=\displaystyle\frac\pi2\)

\[\sin\frac A2 = \cos\frac{B + C}2 \qquad \tan\frac A2 = \cot\frac{B + C}2 \]

正余弦和

\[\begin{aligned} \sin A + \sin B + \sin C &= 4\cos A\ \cos B\ \cos C\\ \cos A + \cos B + \cos C &= 1 + 4\cos \frac A2\ \cos \frac B2\ \cos \frac C2 \end{aligned} \]

正余弦平方和

\[\begin{aligned} \sin^2 A + \sin^2 B + \sin^2 C &= 2 + 2\cos A\ \cos B\ \cos C\\ \cos^2 A + \cos^2 B + \cos^2 C &= 1- 2\cos A\ \cos B\ \cos C \end{aligned} \]

正余弦半角平方和

\[\begin{aligned} \sin^2 \frac A2 + \sin^2 \frac B2 + \sin^2 \frac C2 &= 1 - 2\cos \frac A2\ \cos \frac B2\ \cos \frac C2\\ \cos^2 \frac A2 + \cos^2 \frac B2 + \cos^2 \frac C2 &= 2 + 2\sin \frac A2\ \sin \frac B2\ \sin \frac C2 \end{aligned} \]

正余切和积

\[\tan A + \tan B + \tan C = \tan A\ \tan B\ \tan C\\ \cot \frac A2 + \cot \frac B2 + \cot \frac C2 = \cot \frac A2\ \cot \frac B2\ \cot \frac C2 \]

n倍角公式

切比雪夫多项式 \(g_{n+1}(x)=2x\ g_n(x)-g_{n-1}(x)\) 结论

\[\begin{aligned} \cos 2\alpha &= 2\cos^2\alpha - 1\\ \cos 3\alpha &= 4\cos^3\alpha - 3\cos\alpha\\ \cos 4\alpha &= 8\cos^4\alpha - 8\cos^2\alpha + 1\\ \cos 5\alpha &= 16\cos^5\alpha - 20\cos^3\alpha + 5\cos\alpha\\ \end{aligned} \]

正弦定理

\[\frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = 2R \]

三角形面积公式

\[\begin{aligned} S_\Delta &= \frac12\,ab\,\sin C = \frac12\,ac\,\sin B = \frac12\,bc\,\sin A\\ S_\Delta &= \sqrt{p\,(p-a)(p-b)(p-c)} & &p=\frac{a+b+c}2\\ S_\Delta &= \sqrt{\frac14\left[a^2\,c^2-(\frac{a^2+b^2-c^2}2)^2\right]}\\ S_\Delta &= \frac{abc}{4R} = 2R^2\,\sin A\,\sin B\,\sin C & R&为三角形外接圆半径\\ S_\Delta &= \frac12\,r\,(a+b+c) & r&为三角形内切圆半径\\ \end{aligned} \]

余弦定理

\[\begin{aligned} a^2=b^2+c^2-2\,bc\cos A & & \cos A=\frac{b^2+c^2-a^2}{2bc}\\ b^2=a^2+c^2-2\,ac\cos B & \qquad \Leftrightarrow & \cos B=\frac{a^2+c^2-b^2}{2ac}\\ c^2=a^2+b^2-2\,ab\cos C & & \cos C=\frac{a^2+b^2-c^2}{2ab}\\ \end{aligned} \]

三角形形状判断

\[\cos A \,\cos B \,\cos C < 0 \quad\Leftrightarrow\quad 钝角 \quad\Leftrightarrow\quad \tan A\, \tan B < 1\\ \cos A \,\cos B \,\cos C = 0 \quad\Leftrightarrow\quad 直角 \quad\Leftrightarrow\quad \tan A\, \tan B = 1\\ \cos A \,\cos B \,\cos C > 0 \quad\Leftrightarrow\quad 锐角 \quad\Leftrightarrow\quad \tan A\, \tan B > 1\\ \]

posted @ 2022-03-04 19:10  Howardzhangdqs  阅读(399)  评论(0编辑  收藏  举报