烦人三角函数公式整理

弧、面积公式#

\begin{aligned} l = α \cdot r S = \frac{1}{2} α r^{2} = \frac{1}{2} l r \end{aligned}

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

升降幂公式#

(\sin α + \cos α)^{2} = 1 + 2 \sin α \cos α (\sin α - \cos α)^{2} = 1 - 2 \sin α \cos α

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

诱导公式#

\begin{aligned} \sin (2 k π + α) & = \sin α & \cos (2 k π + α) & = \cos α & \tan (2 k π + α) & = \tan α \end{aligned}

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

\begin{aligned} \sin (π - α) & = \sin α & \cos (π - α) & = - \cos α & \tan (π - α) & = - \tan α \\ \sin (- α) & = - \sin α & \cos (- α) & = \cos α & \tan (- α) & = - \tan α \\ \sin (π + α) & = - \sin α & \cos (π + α) & = - \cos α & \tan (π + α) & = \tan α \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{π}{2} - α) & = \cos α & \sin (\frac{π}{2} + α) & = \cos α \\ \cos (\frac{π}{2} - α) & = \sin α & \cos (\frac{π}{2} + α) & = - \sin α \\ \tan (\frac{π}{2} - α) & = \cot α & \tan (\frac{π}{2} + α) & = - \cot α \\ \cot (\frac{π}{2} - α) & = \tan α & \cot (\frac{π}{2} + α) & = - \tan α \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 π}{2} - α) & = - \cos α & \sin (\frac{3 π}{2} + α) & = - \cos α \\ \cos (\frac{3 π}{2} - α) & = - \sin α & \cos (\frac{3 π}{2} + α) & = \sin α \\ \tan (\frac{3 π}{2} - α) & = - \cot α & \tan (\frac{3 π}{2} + α) & = - \cot α \\ \cot (\frac{3 π}{2} - α) & = \tan α & \cot (\frac{3 π}{2} + α) & = - \tan α \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 (α + β) & = \sin α \cos β + \cos α \sin β & \sin (α - β) & = \sin α \cos β - \cos α \sin β \\ \cos (α + β) & = \cos α \cos β - \sin α \sin β & \cos (α - β) & = \cos α \cos β + \sin α \sin β \\ \tan (α + β) & = \frac{\tan α + \tan β}{1 - \tan α \tan β} & \tan (α - β) & = \frac{\tan α - \tan β}{1 + \tan α \tan β} \end{aligned}

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

倍角公式#

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

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

半角公式#

\begin{aligned} \sin \frac{α}{2} & = \pm \sqrt{\frac{1 - \cos α}{2}} \\ \cos \frac{α}{2} & = \pm \sqrt{\frac{1 + \cos α}{2}} \\ \tan \frac{α}{2} & = \pm \sqrt{\frac{1 - \cos α}{1 + \cos α}} \\ \tan \frac{α}{2} & = \frac{\sin α}{1 + \cos α} = \frac{1 - \cos α}{\sin α} \end{aligned}

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

积化和差#

\begin{aligned} \sin α \cos β & = \frac{1}{2} [\sin (α + β) + \sin (α - β)] & \cos α \sin β & = \frac{1}{2} [\sin (α + β) - \sin (α - β)] \\ \cos α \cos β & = \frac{1}{2} [\cos (α + β) + \cos (α - β)] & \sin α \sin β & = - \frac{1}{2} [\cos (α + β) - \cos (α - β)] \end{aligned}

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

和差化积#

\begin{aligned} \sin α + \sin β & = 2 \sin \frac{α + β}{2} \sin \frac{α - β}{2} & \sin α - \sin β & = 2 \sin \frac{α + β}{2} \sin \frac{α - β}{2} \\ \cos α + \cos β & = 2 \cos \frac{α + β}{2} \cos \frac{α - β}{2} & \cos α - \cos β & = - 2 \cos \frac{α + β}{2} \cos \frac{α - β}{2} \end{aligned}

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

辅助角公式#

a \sin α + b \cos α = \sqrt{a^{2} + b^{2}} \sin (α + φ) (\cos φ = \frac{a}{\sqrt{a^{2} + b^{2}}}, \sin φ = \frac{b}{\sqrt{a^{2} + b^{2}}}, \tan φ = \frac{b}{a})

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

三倍角公式#

\begin{aligned} \sin 3 α & = 3 \sin α - 4 \sin^{3} α = 4 \sin α \sin (\frac{π}{3} - α) \sin (\frac{π}{3} + α) \\ \cos 3 α & = 4 \cos^{3} α - 3 \cos α = 4 \cos α \cos (\frac{π}{3} - α) \cos (\frac{π}{3} + α) \\ \cos 3 α & = \frac{3 \tan α - \tan^{3} α}{1 - 3 \tan^{2} α} = 4 \tan α \tan (\frac{π}{3} - α) \tan (\frac{π}{3} + α) \end{aligned}

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

正余切和差#

\begin{aligned} \tan α + \cot α & = \frac{1}{\sin α \cos α} = \frac{2}{\sin 2 α} \\ \tan α - \cot α & = \frac{\sin^{2} α - \cos^{2} α}{\sin α \cos α} = \frac{- 2 \cos 2 α}{\sin 2 α} = - 2 \cot 2 α \end{aligned}

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

万能公式#

\begin{aligned} \sin α & = \frac{2 \tan \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} & \cos α & = \frac{1 - \tan^{2} \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} & \tan α & = \frac{2 \tan \frac{α}{2}}{1 - \tan^{2} \frac{α}{2}} \end{aligned}

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

降幂公式#

\begin{aligned} \sin^{2} α & = \frac{1 - \cos 2 α}{2} & \cos^{2} α & = \frac{1 + \cos 2 α}{2} & \sin α \cos α & = \frac{\sin 2 α}{2} \end{aligned}

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

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{A}{2} = \cos \frac{B + C}{2} \tan \frac{A}{2} = \cot \frac{B + C}{2}

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

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{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \end{aligned}

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

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}

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

正余弦半角平方和#

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

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

正余切和积#

\tan A + \tan B + \tan C = \tan A \tan B \tan C \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}

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

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 α & = 2 \cos^{2} α - 1 \\ \cos 3 α & = 4 \cos^{3} α - 3 \cos α \\ \cos 4 α & = 8 \cos^{4} α - 8 \cos^{2} α + 1 \\ \cos 5 α & = 16 \cos^{5} α - 20 \cos^{3} α + 5 \cos α \end{aligned}

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

正弦定理#

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

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

三角形面积公式#

\begin{aligned} S_{Δ} & = \frac{1}{2} a b \sin C = \frac{1}{2} a c \sin B = \frac{1}{2} b c \sin A \\ S_{Δ} & = \sqrt{p (p - a) (p - b) (p - c)} & p = \frac{a + b + c}{2} \\ S_{Δ} & = \sqrt{\frac{1}{4} [a^{2} c^{2} - (\frac{a^{2} + b^{2} - c^{2}}{2})^{2}]} \\ S_{Δ} & = \frac{a b c}{4 R} = 2 R^{2} \sin A \sin B \sin C & R & 为 三 角 形 外 接 圆 半 径 \\ S_{Δ} & = \frac{1}{2} r (a + b + c) & r & 为 三 角 形 内 切 圆 半 径 \end{aligned}

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

余弦定理#

\begin{aligned} a^{2} = b^{2} + c^{2} - 2 b c \cos A & \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \\ b^{2} = a^{2} + c^{2} - 2 a c \cos B & \Leftrightarrow & \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} \\ c^{2} = a^{2} + b^{2} - 2 a b \cos C & \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \end{aligned}

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

三角形形状判断#

\cos A \cos B \cos C < 0 \Leftrightarrow 钝 角 \Leftrightarrow \tan A \tan B < 1 \cos A \cos B \cos C = 0 \Leftrightarrow 直 角 \Leftrightarrow \tan A \tan B = 1 \cos A \cos B \cos C > 0 \Leftrightarrow 锐 角 \Leftrightarrow \tan A \tan B > 1

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

posted @ 2022-03-04 19:10 Howardzhangdqs 阅读(483) 评论(0) 编辑收藏举报

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