三角函数之诱导公式

First

如上图:

$$\begin{eqnarray} 已知: \enspace AC=\sin\alpha,BC=\cos\alpha \\ \\ \\ \sin\left(\frac{\pi}{2}-\alpha\right)=\frac{AD}{AB}=\frac{BC}{AB}=\cos\alpha \\ \\ 诱导公式组0.1: \enspace \sin\left(\frac{\pi}{2}-\alpha\right)=\cos\alpha \\ \\ \\ \cos\left(\frac{\pi}{2}-\alpha\right)=\frac{BD}{AB}=\frac{AC}{AB}=\sin\alpha \\ \\ 诱导公式组0.2: \enspace \cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha \\ \\ \\ \tan\left(\frac{\pi}{2}-\alpha\right)=\frac{AD}{BD}=\frac{BC}{AC}=\frac{\cos\alpha}{\sin\alpha} \\ \\ \because \tan\alpha=\frac{\sin\alpha}{\cos\alpha} \\ \\ 诱导公式组0.3: \enspace \tan\left(\frac{\pi}{2}-\alpha\right)=\frac{1}{\tan\alpha}=\cot\alpha \\ \\ \\ 诱导公式0.1: \enspace \sin\left(\frac{\pi}{2}-\alpha\right)=\cos\alpha \\ 诱导公式0.2: \enspace \cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha \\ 诱导公式0.3: \enspace \tan\left(\frac{\pi}{2}-\alpha\right)=\frac{1}{\tan\alpha}=\cot\alpha \end{eqnarray}$$

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Second: 奇偶性质

如上图:

$$\begin{align} \triangle ABC=\triangle ABD \\ \angle BAC=\alpha,\angle DAB=-\alpha \\ \angle BAC=\angle DAB \\ \\ 在 \triangle ABD \\ \sin(-\alpha)=\frac{BD}{AD}=-\sin\alpha \\ \therefore\sin(-\alpha)=-\sin\alpha \\ \\ \cos(-\alpha)=\frac{AB}{AD}=\cos\alpha \\ \therefore\cos(-\alpha)=\cos\alpha \\ \\ 诱导公式1.0(奇函数):\enspace \sin(-\alpha)=-\sin\alpha \\ 诱导公式1.1(偶函数):\enspace \cos(-\alpha)=\cos\alpha \end{align}$$

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Third

$$\begin{align} 标三零:\\ \sin(\frac{\pi}{2}+\alpha)\Rightarrow\sin[\frac{\pi}{2}-(-a)] \\ 根据公式0.1: \\ \sin\left[\frac{\pi}{2}-\left(-a\right)\right]=\cos\left(-a\right) \\ \\ 根据公式1.1: \\ \sin[\frac{\pi}{2}-(-a)]=\cos\alpha \\ \therefore\sin\left(\frac{\pi}{2}+\alpha\right)=\cos\alpha \\ \\ \\ 标三一:\\ \cos(\frac{\pi}{2}+\alpha)\Rightarrow\cos[\frac{\pi}{2}-(-\alpha)] \\ 根据公式0.2: \\ \cos[\frac{\pi}{2}-(-\alpha)]=\sin(-\alpha) \\ \\ 根据公式1.0: \\ \cos[\frac{\pi}{2}-(-\alpha)]=-\sin\alpha \\ \therefore\cos(\frac{\pi}{2}+\alpha)=-\sin\alpha \\ \\ \\ 诱导公式2.0:\enspace \sin\left(\frac{\pi}{2}+\alpha\right)=\cos\alpha \\ 诱导公式2.1:\enspace \cos\left(\frac{\pi}{2}+\alpha\right)=-\sin\alpha \end{align}$$

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Fourth

$$\begin{align} 求证: \quad 1+\tan^{2}x=\sec^{2}x \\ \\ \Rightarrow 1+\frac{\sin^{2}x}{\cos^{2}x}=\frac{\cos^{2}x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x} =\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x} \\ \\ \because \cos^{2}x+\sin^{2}x=1 \\ \\ \therefore \frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}=\frac{1}{\cos^{2}x}=(\frac{1}{\cos x})^{2} \\ \\ \because \frac{1}{\cos x}=\sec x \\ \\ \therefore(\frac{1}{\cos x})^{2}=\sec^{2}x \\ \\ \therefore 1+\tan^{2} x=\sec^{2}x \end{align}$$

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