Remove ads正切半角公式又称万能公式,这一组公式有四个功能: 将角统一为 α 2 {\displaystyle {\frac {\alpha }{2}}} [1]; 将函数名称统一为 tan {\displaystyle \tan } ; 任意实数都可以 tan α 2 {\displaystyle \tan {\frac {\alpha }{2}}} 的形式表达,可用正切函数换元。 在某些积分中,可以将含有三角函数的积分变为有理分式的积分。 因此,这组公式被称为以切表弦公式,简称以切表弦。它们是由二倍角公式求得的。 sin α = 2 tan α 2 1 + tan 2 α 2 {\displaystyle \sin \alpha ={\frac {2\tan {\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}} cos α = 1 − tan 2 α 2 1 + tan 2 α 2 {\displaystyle \cos \alpha ={\frac {1-\tan ^{2}{\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}} tan α = 2 tan α 2 1 − tan 2 α 2 {\displaystyle \tan \alpha ={\frac {2\tan {\frac {\alpha }{2}}}{1-\tan ^{2}{\frac {\alpha }{2}}}}} cot α = 1 − tan 2 α 2 2 tan α 2 {\displaystyle \cot \alpha ={\frac {1-\tan ^{2}{\frac {\alpha }{2}}}{2\tan {\frac {\alpha }{2}}}}} sec α = 1 + tan 2 α 2 1 − tan 2 α 2 {\displaystyle \sec \alpha ={\frac {1+\tan ^{2}{\frac {\alpha }{2}}}{1-\tan ^{2}{\frac {\alpha }{2}}}}} csc α = 1 + tan 2 α 2 2 tan α 2 {\displaystyle \csc \alpha ={\frac {1+\tan ^{2}{\frac {\alpha }{2}}}{2\tan {\frac {\alpha }{2}}}}} 而被称为万能公式的原因是利用 tan α 2 {\displaystyle \tan {\frac {\alpha }{2}}} 的代换可以解决一些有关三角函数的积分。参见三角换元法。 tan ( η 2 ± θ 2 ) = sin η ± sin θ cos η + cos θ = − cos η − cos θ sin η ∓ sin θ , tan ( ± θ 2 ) = ± sin θ 1 + cos θ = ± tan θ sec θ + 1 = ± 1 csc θ + cot θ , ( η = 0 ) tan ( ± θ 2 ) = 1 − cos θ ± sin θ = sec θ − 1 ± tan θ = ± ( csc θ − cot θ ) , ( η = 0 ) tan ( π 4 ± θ 2 ) = 1 ± sin θ cos θ = sec θ ± tan θ = csc θ ± 1 cot θ , ( η = π 2 ) tan ( π 4 ± θ 2 ) = cos θ 1 ∓ sin θ = 1 sec θ ∓ tan θ = cot θ csc θ ∓ 1 , ( η = π 2 ) 1 − tan θ 2 1 + tan θ 2 = 1 − sin θ 1 + sin θ . {\displaystyle {\begin{aligned}\tan \left({\frac {\eta }{2}}\pm {\frac {\theta }{2}}\right)&={\frac {\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }}=-{\frac {\cos \eta -\cos \theta }{\sin \eta \mp \sin \theta }},\\[10pt]\tan \left(\pm {\frac {\theta }{2}}\right)&={\frac {\pm \sin \theta }{1+\cos \theta }}={\frac {\pm \tan \theta }{\sec \theta +1}}={\frac {\pm 1}{\csc \theta +\cot \theta }},~~~~(\eta =0)\\[10pt]\tan \left(\pm {\frac {\theta }{2}}\right)&={\frac {1-\cos \theta }{\pm \sin \theta }}={\frac {\sec \theta -1}{\pm \tan \theta }}=\pm (\csc \theta -\cot \theta ),~~~~(\eta =0)\\[10pt]\tan \left({\frac {\pi }{4}}\pm {\frac {\theta }{2}}\right)&={\frac {1\pm \sin \theta }{\cos \theta }}=\sec \theta \pm \tan \theta ={\frac {\csc \theta \pm 1}{\cot \theta }},~~~~(\eta ={\frac {\pi }{2}})\\[10pt]\tan \left({\frac {\pi }{4}}\pm {\frac {\theta }{2}}\right)&={\frac {\cos \theta }{1\mp \sin \theta }}={\frac {1}{\sec \theta \mp \tan \theta }}={\frac {\cot \theta }{\csc \theta \mp 1}},~~~~(\eta ={\frac {\pi }{2}})\\[10pt]{\frac {1-\tan {\frac {\theta }{2}}}{1+\tan {\frac {\theta }{2}}}}&={\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}.\end{aligned}}} Remove ads万能公式的证明 由二倍角公式,有: sin α = 2 sin α 2 cos α 2 = 2 sin α 2 cos α 2 cos 2 α 2 + sin 2 α 2 = 2 sin α 2 cos α 2 ÷ cos 2 α 2 ( cos 2 α 2 + sin 2 α 2 ) ÷ cos 2 α 2 = 2 sin α 2 cos α 2 1 + sin 2 α 2 cos 2 α 2 = 2 tan α 2 1 + tan 2 α 2 {\displaystyle \sin \alpha =2\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}={\frac {2\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\alpha }{2}}}}={\frac {2\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}\div \cos ^{2}{\frac {\alpha }{2}}}{(\cos ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\alpha }{2}})\div \cos ^{2}{\frac {\alpha }{2}}}}={\frac {2{\frac {\sin {\frac {\alpha }{2}}}{\cos {\frac {\alpha }{2}}}}}{1+{\frac {\sin ^{2}{\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}}}}}={\frac {2\tan {\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}} tan α = 2 tan α 2 1 − tan 2 α 2 {\displaystyle \tan \alpha ={\frac {2\tan {\frac {\alpha }{2}}}{1-\tan ^{2}{\frac {\alpha }{2}}}}} 再由同角三角函数间的关系,得出 cos α = sin α tan α = 2 tan α 2 1 + tan 2 α 2 2 tan α 2 1 − tan 2 α 2 = 1 − tan 2 α 2 1 + tan 2 α 2 {\displaystyle \cos \alpha ={\frac {\sin \alpha }{\tan \alpha }}={\frac {\frac {2\tan {\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}{\frac {2\tan {\frac {\alpha }{2}}}{1-\tan ^{2}{\frac {\alpha }{2}}}}}={\frac {1-\tan ^{2}{\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}} Remove ads几何证明 正切半角公式的几何证明 在单位圆内, t = tan ϕ 2 {\displaystyle t=\tan {\frac {\phi }{2}}} 。根据相似关系, t sin ϕ = 1 1 + cos ϕ {\displaystyle {\frac {t}{\sin \phi }}={\frac {1}{1+\cos \phi }}} ,可得出 t = sin ϕ 1 + cos ϕ = sin ϕ ( 1 − cos ϕ ) ( 1 + cos ϕ ) ( 1 − cos ϕ ) = 1 − cos ϕ sin ϕ {\displaystyle t={\frac {\sin \phi }{1+\cos \phi }}={\frac {\sin \phi (1-\cos \phi )}{(1+\cos \phi )(1-\cos \phi )}}={\frac {1-\cos \phi }{\sin \phi }}} 。 显然 tan a + b 2 = sin a + b 2 cos a + b 2 = sin a + sin b cos a + cos b {\displaystyle \tan {\frac {a+b}{2}}={\frac {\sin {\frac {a+b}{2}}}{\cos {\frac {a+b}{2}}}}={\frac {\sin a+\sin b}{\cos a+\cos b}}} 。 Remove ads双曲函数 此公式亦可以对双曲函数起到类似的作用,由双曲线右支上的一点 ( cosh θ , sinh θ ) {\displaystyle (\cosh \theta ,\sinh \theta )} 给出。从 ( − 1 , 0 ) {\displaystyle (-1,0)} 到y轴给出了如下等式: t = tanh 1 2 θ = sinh θ cosh θ + 1 = cosh θ − 1 sinh θ {\displaystyle t=\tanh {\frac {1}{2}}\theta ={\frac {\sinh \theta }{\cosh \theta +1}}={\frac {\cosh \theta -1}{\sinh \theta }}} 可以得到 cosh θ = 1 + t 2 1 − t 2 , {\displaystyle \cosh \theta ={\frac {1+t^{2}}{1-t^{2}}},} sinh θ = 2 t 1 − t 2 , {\displaystyle \sinh \theta ={\frac {2t}{1-t^{2}}},} tanh θ = 2 t 1 + t 2 , {\displaystyle \tanh \theta ={\frac {2t}{1+t^{2}}},} coth θ = 1 + t 2 2 t , {\displaystyle \coth \theta ={\frac {1+t^{2}}{2t}},} s e c h θ = 1 − t 2 1 + t 2 , {\displaystyle \mathrm {sech} \,\theta ={\frac {1-t^{2}}{1+t^{2}}},} c s c h θ = 1 − t 2 2 t , {\displaystyle \mathrm {csch} \,\theta ={\frac {1-t^{2}}{2t}},} 和 e θ = 1 + t 1 − t , {\displaystyle e^{\theta }={\frac {1+t}{1-t}},} e − θ = 1 − t 1 + t . {\displaystyle e^{-\theta }={\frac {1-t}{1+t}}.} 卡尔·维尔斯特拉斯引入这个式子来省去查找原函数的麻烦。 θ {\displaystyle \theta } 在 T {\displaystyle T} 而得出下面的双曲反正切函数和自然对数之间的关系: artanh t = 1 2 ln 1 + t 1 − t . {\displaystyle \operatorname {artanh} t={\frac {1}{2}}\ln {\frac {1+t}{1-t}}.} Remove ads参见 三角恒等式 参考文献Loading content...外部链接Loading content...Loading related searches...Wikiwand - on Seamless Wikipedia browsing. 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