三角小题

链剑心

设$x,y,z \in R,x+y+z=\pi$
\[\tan\left(\frac{x+y-z}{4}\right)+\tan\left(\frac{y+z-x}{4}\right)+\tan\left(\frac{z+x-y}{4}\right)=1\]
求证：
\[\cos x+\cos y+\cos z=1\]

kuing

令 $a=\tan (x/2)$, $b=\tan (y/2)$, $c=\tan (z/2)$，则由 $x+y+z=\pi$ 知 $ab+bc+ca=1$，于是
\begin{align*}
\sum\tan \frac{x+y-z}4&=\sum\tan \left( \frac\pi4-\frac z2 \right) \\
& =\sum\frac{1-c}{1+c} \\
& =\frac{3+a+b+c-ab-bc-ca-3abc}{1+a+b+c+ab+bc+ca+abc} \\
& =\frac{2+a+b+c-3abc}{2+a+b+c+abc},
\end{align*}
所以只能 $abc=0$，即
\[\sin \frac x2\sin \frac y2\sin \frac z2=0,\]
由积化和差，有
\begin{align*}
\sin \frac x2\sin \frac y2\sin \frac z2&=\frac12\cos \frac{x-y}2\sin \frac z2-\frac12\cos \frac{x+y}2\sin \frac z2 \\
& =\frac14\sin \frac{x-y+z}2-\frac14\sin \frac{x-y-z}2-\frac14\sin \frac{x+y+z}2+\frac14\sin \frac{x+y-z}2 \\
& =\frac14\cos y+\frac14\cos x-\frac14+\frac14\cos z,
\end{align*}
所以
\[\cos x+\cos y+\cos z=1.\]

链剑心

回复 2# kuing

ab+bc+ca=1是为什么

那个貌似在三角形中才成立吧

kuing

回复 3# 链剑心

不在三角形内也是成立的

链剑心

回复 4# kuing


    说说看

kuing

$x+y+z=(2k+1)\pi$, $x$, $y$, $z\in\Bbb R$, $k\in\Bbb Z$，则
\begin{align*}
& \tan \frac x2\tan \frac y2+\tan \frac y2\tan \frac z2+\tan \frac z2\tan \frac x2 \\
={}&\tan \frac x2\tan \frac y2+\left( \tan \frac x2+\tan \frac y2 \right)\tan\left(k\pi+\frac\pi2-\frac{x+y}2\right) \\
={}&\tan \frac x2\tan \frac y2+\left( \tan \frac x2+\tan \frac y2 \right)\cot \frac{x+y}2 \\
={}&\tan \frac x2\tan \frac y2+\left( \tan \frac x2+\tan \frac y2 \right)\frac{1-\tan \frac x2\tan \frac y2}{\tan \frac x2+\tan \frac y2} \\
={}&1 .
\end{align*}

当然了，上面的过程（包括二楼）可能还有一些细节未处理，就是可能涉及一些让 tan 无意义的情况未曾讨论，时间关系就懒了，反正至少可以肯定非特殊角的情况是没问题的……

其妙

两个著名的三角正切恒等式之一

其妙

这里也有：blog.sina.com.cn/s/blog_4c1131020101esad.html

hbghlyj

Further "conditional" identities for the case $α+β+γ=180°$
The following formulae apply to arbitrary plane triangles and follow from $\alpha + \beta + \gamma = 180^{\circ}$, as long as the functions occurring in the formulae are well-defined (the latter applies only to the formulae in which tangents and cotangents occur).
\begin{align}
   \tan \alpha + \tan \beta + \tan \gamma &= \tan \alpha \tan \beta \tan \gamma \\
   1 &= \cot \beta \cot \gamma + \cot \gamma \cot \alpha + \cot \alpha \cot \beta \\
   \cot\left(\frac{\alpha}{2}\right) + \cot\left(\frac{\beta}{2}\right) + \cot\left(\frac{\gamma}{2}\right) &= \cot\left(\frac{\alpha}{2}\right) \cot \left(\frac{\beta}{2}\right) \cot\left(\frac{\gamma}{2}\right) \\
   1 &= \tan\left(\frac{\beta}{2}\right)\tan\left(\frac{\gamma}{2}\right) + \tan\left(\frac{\gamma}{2}\right)\tan\left(\frac{\alpha}{2}\right) + \tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right) \\
   \sin \alpha + \sin \beta + \sin \gamma &= 4\cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right)\cos\left(\frac{\gamma}{2}\right) \\
  -\sin \alpha + \sin \beta + \sin \gamma &= 4\cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\sin\left(\frac{\gamma}{2}\right) \\
   \cos \alpha + \cos \beta + \cos \gamma &= 4\sin\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\sin \left(\frac{\gamma}{2}\right) + 1 \\
  -\cos \alpha + \cos \beta + \cos \gamma &= 4\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right)\cos \left(\frac{\gamma}{2}\right) - 1 \\
   \sin (2\alpha) + \sin (2\beta) + \sin (2\gamma) &=  4\sin \alpha \sin \beta \sin \gamma \\
  -\sin (2\alpha) + \sin (2\beta) + \sin (2\gamma) &=  4\sin \alpha \cos \beta \cos \gamma \\
   \cos (2\alpha) + \cos (2\beta) + \cos (2\gamma) &= -4\cos \alpha \cos \beta \cos \gamma - 1 \\
  -\cos (2\alpha) + \cos (2\beta) + \cos (2\gamma) &= -4\cos \alpha \sin \beta \sin \gamma + 1 \\
   \sin^2\alpha + \sin^2\beta + \sin^2\gamma &=  2 \cos \alpha \cos \beta \cos \gamma + 2 \\
  -\sin^2\alpha + \sin^2\beta + \sin^2\gamma &=  2 \cos \alpha \sin \beta \sin \gamma \\
   \cos^2\alpha + \cos^2\beta + \cos^2\gamma &= -2 \cos \alpha \cos \beta \cos \gamma + 1 \\
  -\cos^2\alpha + \cos^2\beta + \cos^2\gamma &= -2 \cos \alpha \sin \beta \sin \gamma + 1 \\
  \sin^2 (2\alpha) + \sin^2 (2\beta) + \sin^2 (2\gamma) &= -2\cos (2\alpha) \cos (2\beta) \cos (2\gamma)+2 \\
  \cos^2 (2\alpha) + \cos^2 (2\beta) + \cos^2 (2\gamma) &=  2\cos (2\alpha) \,\cos (2\beta) \,\cos (2\gamma) + 1 \\
   1 &= \sin^2 \left(\frac{\alpha}{2}\right) + \sin^2 \left(\frac{\beta}{2}\right) + \sin^2 \left(\frac{\gamma}{2}\right) + 2\sin \left(\frac{\alpha}{2}\right) \,\sin \left(\frac{\beta}{2}\right) \,\sin \left(\frac{\gamma}{2}\right)
\end{align}