「学习笔记」三角函数

有关三角函数的公式、转化、证明等等

\[\sin^2 \alpha + \cos^2 \alpha = 1\\ \dfrac{\sin \alpha}{\cos \alpha} = \tan \alpha\\ (\sin \alpha \pm \cos \alpha)^2 = 1 \pm 2 \sin \alpha \cos \alpha\\ \sin \alpha = \cos \alpha \tan \alpha\\ \sin^2 \alpha = \dfrac{\sin^2 \alpha}{\sin^2 \alpha + \cos^2 \alpha} = \dfrac{\tan^2 \alpha}{\tan^2 \alpha + 1}\\ \cos^2 \alpha = \dfrac{\cos^2 \alpha}{\sin^2 \alpha + \cos^2 \alpha} = \dfrac{1}{\tan^2 \alpha + 1}\\ \]

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\[\sin (\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta\\ \cos (\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \cos \beta\\ \tan (\alpha \pm \beta) = \dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}\\ \]

\(\tan (\alpha \pm \beta) = \dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}\) 的证明：

\[\tan (\alpha \pm \beta) = \dfrac{\sin (\alpha \pm \beta)}{\cos (\alpha \pm \beta)} = \dfrac{\sin \alpha \cos \beta \pm \cos \alpha \sin \beta}{\cos \alpha \cos \beta \mp \sin \alpha \cos \beta}\\ 分子分母同除 \cos \alpha \cos \beta\\ \dfrac{\sin \alpha \cos \beta \pm \cos \alpha \sin \beta}{\cos \alpha \cos \beta \mp \sin \alpha \cos \beta} = \dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}\\ \]

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

\(\tan 2 \alpha = \dfrac{2 \tan \alpha}{1 - \tan^2 \alpha}\) 的证明：

\[\tan 2 \alpha = \dfrac{\sin 2 \alpha}{\cos 2 \alpha} = \dfrac{2 \sin \alpha \cos \alpha}{\cos^2 \alpha - \sin^2 \alpha}\\ 分子分母同除 \cos^2 \alpha\\ \tan 2 \alpha = \dfrac{2 \tan \alpha}{1 - \tan^2 \alpha} \]

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\[a \sin \alpha + b \cos \alpha = \sqrt{a^2 + b^2} \sin (\alpha + \varphi)\\ \cos \varphi = \dfrac{a}{\sqrt{a^2 + b^2}}, \sin \varphi = \dfrac{b}{\sqrt{a^2 + b^2}} \]

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\[\tan \alpha \pm \tan \beta = \tan(\alpha \pm \beta)(1 \mp \tan \alpha \tan \beta) \]

证明：

\[\tan \alpha \pm \tan \beta = \dfrac{\sin \alpha}{\cos \alpha} \pm \dfrac{\sin \beta}{\cos \beta} = \dfrac{\sin (\alpha \pm \beta)}{\cos \alpha \cos \beta}\\ 分子分母同除 \cos (\alpha \pm \beta)\\ \dfrac{\sin (\alpha \pm \beta)}{\cos \alpha \cos \beta} = \dfrac{\tan (\alpha \pm \beta)}{\dfrac{\cos \alpha \cos \beta}{\cos \alpha \cos \beta \mp \sin \alpha \sin \beta}} = \dfrac{\tan (\alpha \pm \beta){(\cos \alpha \cos \beta \mp \sin \alpha \sin \beta)}}{\cos \alpha \cos \beta} = \tan (\alpha \pm \beta)(1 \mp \tan \alpha \tan \beta) \]

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\[1 + \cos \alpha = 1 + \cos (\dfrac{\alpha}{2} + \dfrac{\alpha}{2}) = 1 + \cos ^ 2 \dfrac{\alpha}{2} - \sin ^ 2 \dfrac{\alpha}{2} = 2 \cos^2 \dfrac{\alpha}{2}\\ 1 - \cos \alpha = 1 - \cos (\dfrac{\alpha}{2} + \dfrac{\alpha}{2}) = 1 - \cos^2 \dfrac{\alpha}{2} + \sin^2 \dfrac{\alpha}{2} = 2 \sin^2 \dfrac{\alpha}{2}\\ \dfrac{1 + \cos 2 \alpha}{2} = \cos^2 \alpha\\ \dfrac{1 - \cos 2 \alpha}{2} = \sin^2 \alpha\\ \sin 2 \alpha = \dfrac{2 \sin \alpha \cos \alpha}{\sin^2 \alpha + \cos^2 \alpha} = \dfrac{2 \tan \alpha}{\tan^2 \alpha + 1}\\ \cos 2 \alpha = \dfrac{\cos^2 \alpha - \sin^2 \alpha}{\sin^2 \alpha + \cos^2 \alpha} = \dfrac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha}\\ 1 \pm \sin \alpha = \sin^2 \dfrac{\alpha}{2} + \cos^2 \dfrac{\alpha}{2} \pm 2 \sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} = (\sin \dfrac{\alpha}{2} \pm \cos \dfrac{\alpha}{2})^2\\ \tan \dfrac{\alpha}{2} = \dfrac{\sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}} = \sqrt{\dfrac{1 - \cos \alpha}{1 + \cos \alpha}} = \dfrac{\sin \alpha}{1 + \cos \alpha} = \dfrac{1 - \cos \alpha}{\sin \alpha}\\ \]