浅谈三角函数

初识三角函数

三角函数六边形记忆法则

规律1. 六边形对角线互为倒数（倒数关系）

\(\cos x=\dfrac{1}{\sec x}\) 或者 \(\sec x=\dfrac{1}{\cos x}\)

\(\sin x=\dfrac{1}{c s c x}\) 或者 \(\csc x=\dfrac{1}{\sin x}\)

\(\tan x=\dfrac{1}{\cot x}\) 或者 \(\cot x=\dfrac{1}{\tan x}\)

规律2. 灰色三角形上端的平方之和等于下端的平方 (平方关系)

\(\sin ^{2} x+\cos ^{2} x=1\) \(\tan ^{2} x+1=\sec ^{2} x\) \(1+\cot ^{2} x=\csc ^{2} x\)

规律3. 任意一点的值等于这一点顺时针的第一个值与第二个值的比值

$$\begin{aligned} &\tan x=\dfrac{\sin x}{\cos x} \\\\ &\sin x=\dfrac{\cos x}{\cot x} \\\\ &\cos x=\dfrac{\cot x}{\csc x} \\\\ &\cot x=\dfrac{\csc x}{\sec x} \\\\ &\csc x=\dfrac{\sec x}{\tan x} \\\\ &\sec x=\dfrac{\tan x}{\sin x} \end{aligned}$$

规律4. 任意一点的值等于紧挨着这一点的两个端点的值的积

\[\tan x=\sin x \times \sec x\] \[\sin x=\cos x \times \tan x\]

\(\sin\)函数的各常数

\(y = A\sin (\omega x + \varphi ), x\in[0,+\infty)\),其中\(A>0,\omega>0\).

• \(A\)叫做振幅，表示做简谐运动物体离开平衡位置的最大距离。
• 这个简谐运动的频率\(f=\dfrac{1}{T}=\dfrac{\omega}{2\pi}\)
• \(\omega x+\varphi\)称为相位
• \(x=0\)时的相位\(\varphi\)称为初相

基本性质与图像变换

	\(y=\sin x\)	\(y=\cos x\)	\(y=\tan x\)
奇偶性	奇函数	偶函数	奇函数
单调性	\(\left[ {-\dfrac{\pi}{2} + 2k\pi,\dfrac{\pi }{2} + 2k\pi}\right]\)上递增 \(\left[ {\dfrac{\pi}{2} + 2k\pi,\dfrac{3\pi}{2} + 2k\pi}\right]\)上递碱	\(\left[ -\pi+2k\pi,2k\pi\right]\)上递增 \(\left[2k\pi,\pi+2k\pi\right]\)上递减	\(\left( { - \dfrac{\pi }{2} + k\pi,\dfrac{\pi }{2}+k\pi} \right)\)上递增
周期性	\(y = A\sin (\omega x + \varphi )(\omega \ne 0)\) \(T=\dfrac{2\pi }{\mid \omega \mid }\)	\(y = A\cos(\omega x + \varphi )(\omega \ne 0)\) \(T=\dfrac{2\pi }{\mid \omega \mid }\)	\(y = A\tan (\omega x + \varphi )(\omega \ne 0)\) \(T=\dfrac{\pi }{\mid \omega \mid }\)

图像变换

• 平移变换：左加右减，上加下减
• 周期变换：\(y = \sin x \to y = \sin \omega x\)，横坐标伸长/缩短为原来的\(\dfrac{1}{\omega}\)倍
• 振幅变换：\(y = \sin x \to y = A\sin x\)，纵坐标伸长/缩短为原来的\(A\)倍

三角函数恒等变换

诱导公式

口诀

奇变偶不变，符号看象限。

“符号看象限”的含义是：把角α看做锐角，不考虑α角所在象限，看\(n·(\dfrac{\pi}{2})±α\)是第几象限角，从而得到等式右边是正号还是负号。可结合单位圆判断。

\[\left\{ \begin{array}{l} \sin (2k\pi + \alpha ) = \sin \alpha \\ \cos (2k\pi + \alpha ) = \cos \alpha \\ \tan (2k\pi + \alpha ) = \tan \alpha \\ \cot (2k\pi + \alpha ) = \cot \alpha \end{array} \right.\]

\[\left\{ \begin{array}{l} \sin (\pi + \alpha ) = - \sin \alpha \\ \cos (\pi + \alpha ) = - \cos \alpha \\ \tan (\pi + \alpha ) = \tan \alpha \\ \cot (\pi + \alpha ) = \cot \alpha \end{array} \right.\]

\[\left\{ \begin{array}{l} \sin (\pi - \alpha ) = \sin \alpha \\ \cos (\pi - \alpha ) = - \cos \alpha \\ \tan (\pi - \alpha ) = - \tan \alpha \\ \cot (\pi - \alpha ) = - \cot \alpha \end{array} \right.\]

\[\left\{ \begin{array}{l} \sin \left( {\dfrac{\pi }{2} + \alpha } \right) = \cos \alpha \\ \sin \left( {\dfrac{\pi }{2} - \alpha } \right) = \cos \alpha \\ \cos \left( {\dfrac{\pi }{2} + \alpha } \right) = - \sin \alpha \\ \cos \left( {\dfrac{\pi }{2} - \alpha } \right) = \sin \alpha \\ \tan \left( {\dfrac{\pi }{2} + \alpha } \right) = - \cot \alpha \\ \tan \left( {\dfrac{\pi }{2} - \alpha } \right) = \cot \alpha \\ \cot \left( {\dfrac{\pi }{2} + \alpha } \right) = - \tan \alpha \\ \cot \left( {\dfrac{\pi }{2} - \alpha } \right) = \tan \alpha \end{array} \right.\]

和差角公式

\[\sin \left( {\alpha \pm \beta } \right) = \sin \alpha \cdot \cos \beta \pm \cos \alpha \cdot \sin \beta\] \[\cos \left( {\alpha \pm \beta } \right) = \cos \alpha \cdot \cos \beta \mp \sin \alpha \cdot \sin \beta\]

二倍角公式

$$\sin 2\alpha = 2\sin \alpha \cos \alpha = \dfrac{{2\tan \alpha }}{{1 + {{\tan }^2}\alpha }}\\\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha = 1 - 2{\sin ^2}\alpha = 2{\cos ^2}\alpha - 1\\\tan 2\alpha = \dfrac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha}}$$

半角公式

$$ \sin \dfrac{\alpha }{2} = \pm \sqrt {\dfrac{{1 - \cos \alpha }}{2}} \\ \cos \dfrac{\alpha }{2} = \pm \sqrt {\dfrac{{1 + \cos \alpha }}{2}} \\ \tan \dfrac{\alpha }{2} = \pm \sqrt {\dfrac{{1 - \cos \alpha }}{{1 + \cos \alpha }}} = \dfrac{{\sin \alpha }}{{1 + \cos \alpha }} = \dfrac{{1 - \cos \alpha }}{{\sin \alpha }} $$

二倍角、半角公式可用于降次。

万能公式

\[\sin \alpha = \dfrac{2\tan \dfrac{\alpha}{2}}{1 + \tan ^2 \dfrac{\alpha }{2}}\]

\[\cos \alpha = \dfrac{1 - \tan ^2 \dfrac{\alpha }{2}}{1 + \tan ^2 \dfrac{\alpha }{2}}\]

\[\tan \alpha = \dfrac{2\tan \dfrac{\alpha}{2}}{1 - \tan ^2 \dfrac{\alpha }{2}}\]

积化和差

\[\sin \alpha \cdot \cos \beta = \dfrac{1}{2}\left[ {\sin \left( {\alpha + \beta } \right) + \sin \left( {\alpha - \beta } \right)} \right]\]

\[\cos \alpha \cdot \sin \beta = \dfrac{1}{2}\left[ {\sin \left( {\alpha + \beta } \right) - \sin \left( {\alpha - \beta } \right)} \right]\]

\[\cos \alpha \cdot \cos \beta = \dfrac{1}{2}\left[ {\cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right)} \right]\]

\[\sin \alpha \cdot \sin \beta = - \dfrac{1}{2}\left[ {\cos \left( {\alpha + \beta } \right) - \cos \left( {\alpha - \beta } \right)} \right]\]

口诀

正和正在先，正差正后迁；余和一色余，余差翻了天。

和差化积

\[\sin \alpha + \sin \beta = 2\sin \dfrac{\alpha + \beta }{2} \cdot \cos \dfrac{\alpha - \beta }{2}\]

\[\sin \alpha - \sin \beta = 2\cos \dfrac{\alpha + \beta }{2} \cdot \sin \dfrac{\alpha - \beta }{2}\]

\[\cos \alpha + \cos \beta = 2\cos \dfrac{\alpha + \beta }{2} \cdot \cos \dfrac{\alpha - \beta }{2}\]

\[\cos \alpha - \cos \beta = - 2\sin \dfrac{\alpha + \beta }{2} \cdot \sin \dfrac{\alpha - \beta }{2}\]

口诀

正加正，正在前； 正减正，余在前；
余加余，余并肩； 余减余，负正弦。

$ A (t+u)+B (w t+z)$$= (t-- ) $

其他公式

$$\sin 3 \alpha = 3 \sin \alpha - 4{\sin ^3}\alpha = 4 \cdot \sin (60^\circ - \alpha ) \cdot \sin \alpha \cdot \sin (60^\circ + \alpha)$$ $$\cos 3\alpha = 4{\cos ^3}\alpha - 3\cos \alpha = 4 \cdot \cos (60^\circ - \alpha ) \cdot \cos \alpha \cdot \cos (60^\circ + \alpha )$$ $$\tan 3\alpha = \tan (60^\circ - \alpha ) \cdot \tan \alpha \cdot \tan (60^\circ + \alpha )$$ $${\sin ^2}\alpha - {\sin ^2}\beta = \sin (\alpha + \beta ) \cdot \sin(\alpha - \beta )$$ $${\cos ^2}\alpha - {\cos ^2}\beta = - \sin (\alpha + \beta ) \cdot \sin(\alpha - \beta )$$ 【辅助角公式】$a\sin x + b\cos x = \sqrt {{a^2} + {b^2}} \sin (x + \theta )$,其中$\tan\theta=\dfrac{b}{a}$.