一道三角函数题

hbghlyj

$\left\{\begin{aligned}
\sin \alpha &=\sin (\alpha+\beta+\gamma)+1 \\
\sin \beta &=3 \sin (\alpha+\beta+\gamma)+2 \\
\sin \gamma &=5 \sin (\alpha+\beta+\gamma)+3
\end{aligned}\right.$
求$\sin\alpha\sin\beta\sin\gamma$的所有可能值.

hejoseph

只能有
\[
\sin(\alpha+\beta+\gamma)=-\frac 12
\]
或
\[
\sin(\alpha+\beta+\gamma)=-\frac 34
\]

facebooker

回复 2# hejoseph

何版 给点提示啊 思路啥的

kuing

估计有特殊的巧解，我只会暴算

为方便码代码将 `\alpha`, `\beta`, `\gamma` 改为 `a`, `b`, `c`，设 `x=\sin(a+b+c)`，完全展开有
\[x=\sin a\cos b\cos c+\cos a\sin b\cos c+\cos a\cos b\sin c-\sin a\sin b\sin c,\]移项平方得
\[(x-\cos a\cos b\sin c+\sin a\sin b\sin c)^2=(\sin a\cos b+\cos a\sin b)^2(1-\sin^2c),\]展开得
\begin{align*}
&x^2+(1-\sin^2a)(1-\sin^2b)\sin^2c+\sin^2a\sin^2b\sin^2c\\
&-2x\cos a\cos b\sin c+2x\sin a\sin b\sin c-2\cos a\cos b\sin a\sin b\sin^2c\\
={}&(\sin^2a(1-\sin^2b)+(1-\sin^2a)\sin^2b+2\sin a\cos b\cos a\sin b)(1-\sin^2c),
\end{align*}将含 `\cos a\cos b` 的移到一边并化简得
\begin{align*}
&x^2+2x\sin a\sin b\sin c-\sin^2a-\sin^2b+2\sin^2a\sin^2b+\sin^2c\\
={}&2\cos a\cos b(x\sin c+\sin a\sin b),
\end{align*}再平方得
\begin{align*}
&(x^2+2x\sin a\sin b\sin c-\sin^2a-\sin^2b+2\sin^2a\sin^2b+\sin^2c)^2\\
={}&4(1-\sin^2a)(1-\sin^2b)(x\sin c+\sin a\sin b)^2,
\end{align*}最后代入条件 `\sin a=x+1`, `\sin b=3x+2`, `\sin c=5x+3`，恰好化简为
\[16(2x+1)^4(4x+3)^2=0.\]

hbghlyj

没找到好方法。但是有恒等式：
若$\sum\limits_{i=1}^3\arcsin a_i\equiv0\pmod{2\pi}$,则$s_1^4-4 s_2s_1^2+8 s_3 s_1+4 s_3^2=0$
若$\sum\limits_{i=1}^4\arcsin a_i\equiv0\pmod{2\pi}$,则$s_1^4-4 s_2s_1^2+8 s_3 s_1+4 s_3^2-4 s_4 s_1^2=0$
若$\sum\limits_{i=1}^5\arcsin a_i\equiv0\pmod{2\pi}$,则$\left(s_1^4-4 s_2 s_1^2-4 s_4 s_1^2+8 s_3 s_1+4 s_3^2\right)^2 \left(s_1^4-4 s_2 s_1^2+4 s_4 s_1^2+8 s_3 s_1+4 s_3^2-16 s_2 s_4-16 s_4\right)^2+32 s_5\left(s_1^4-4 s_2 s_1^2-4 s_4 s_1^2+8 s_3 s_1+4 s_3^2\right) \left(s_1^4-4 s_2 s_1^2+4 s_4 s_1^2+8 s_3 s_1+4 s_3^2-16 s_2 s_4-16 s_4\right) \left(s_1^5+2 s_3 s_1^4-4 s_2 s_1^3-8 s_2 s_3 s_1^2+4 s_3^2 s_1-8 s_2 s_4 s_1-8 s_4 s_1+8 s_2^2 s_3+16 s_2 s_3+8 s_3\right)+64s_5^2\left(\left(s_1^4-4 s_2 s_1^2+8 s_3 s_1+4 s_3^2\right) \left(s_1^8-8 s_2 s_1^6-10 s_1^6-4 s_3 s_1^5+34 s_2^2 s_1^4+4 s_3^2 s_1^4+72 s_2 s_1^4+46 s_1^4+16 s_2 s_3 s_1^3+48 s_3 s_1^3-72 s_2^3 s_1^2-224 s_2^2 s_1^2-16 s_2 s_3^2 s_1^2+24 s_3^2 s_1^2-184 s_2 s_1^2-32 s_1^2+16 s_3^3 s_1-48 s_2^2 s_3 s_1+48 s_3 s_1+64 s_2^4+192 s_2^3+192 s_2^2+8 s_2^2 s_3^2+64 s_2 s_3^2+56 s_3^2+64 s_2\right)+4 s_4\left(4 s_2 s_1^8+3 s_1^8-32 s_2^2 s_1^6-64 s_2 s_1^6-8 s_1^6-16 s_2 s_3 s_1^5+32 s_3 s_1^5+88 s_2^3 s_1^4+232 s_2^2 s_1^4-16 s_2 s_3^2 s_1^4+8 s_3^2 s_1^4+72 s_2 s_1^4-88 s_1^4+64 s_2^2 s_3 s_1^3-192 s_2 s_3 s_1^3-320 s_3 s_1^3-96 s_2^4 s_1^2-160 s_2^3 s_1^2+32 s_2^2 s_1^2+64 s_2^2 s_3^2 s_1^2-128 s_2 s_3^2 s_1^2-288 s_3^2 s_1^2+160 s_2 s_1^2+64 s_1^2-64 s_2 s_3^3 s_1-128 s_3^3 s_1+64 s_2^3 s_3 s_1+64 s_2^2 s_3 s_1-64 s_2 s_3 s_1-64 s_3 s_1-16 s_3^4-32 s_2^3 s_3^2-160 s_2^2 s_3^2-224 s_2 s_3^2-96 s_3^2\right)+32 s_1 s_4^2\left(2 s_3 s_1^4+s_2^2 s_1^3-4 s_2 s_1^3-s_1^3-8 s_2 s_3 s_1^2-4 s_2^3 s_1+12 s_2^2 s_1+4 s_3^2 s_1+36 s_2 s_1+20 s_1+16 s_2^2 s_3+32 s_2 s_3+16 s_3\right)-64 s_1^4s_4^3\right)-1024 s_5^3\left(s_1^9-8 s_2 s_1^7-6 s_1^7+4 s_3 s_1^6+22 s_2^2 s_1^5+4 s_3^2 s_1^5+32 s_2 s_1^5+18 s_1^5-4 s_2^2 s_3 s_1^4-24 s_2 s_3 s_1^4-4 s_3 s_1^4-24 s_2^3 s_1^3-48 s_2^2 s_1^3-16 s_2 s_3^2 s_1^3-8 s_3^2 s_1^3-16 s_4^2 s_1^3-72 s_2 s_1^3-48 s_1^3+16 s_2^3 s_3 s_1^2+80 s_2^2 s_3 s_1^2-16 s_2 s_3 s_1^2-80 s_3 s_1^2+32 s_2^3 s_1+96 s_2^2 s_1+24 s_2^2 s_3^2 s_1-64 s_2 s_3^2 s_1-88 s_3^2 s_1+96 s_2 s_1+32 s_1-32 s_2 s_3^3-32 s_3^3-16 s_2^4 s_3-32 s_2^3 s_3+32 s_2 s_3+16 s_3+\left(8 s_2 s_1^5-4 s_1^5-8 s_3 s_1^4-32 s_2^2 s_1^3-32 s_2 s_1^3+16 s_1^3+32 s_2 s_3 s_1^2+64 s_3 s_1^2+16 s_2^3 s_1+80 s_2^2 s_1+16 s_3^2 s_1+112 s_2 s_1+48 s_1+32 s_2^2 s_3+64 s_2 s_3+32 s_3\right) s_4\right)+2048s_5^4 \left(3 s_1^6-14 s_2 s_1^4-16 s_1^4+12 s_2^2 s_1^2+4 s_3^2 s_1^2+40 s_2 s_1^2+8 s_2 s_4 s_1^2-16 s_4 s_1^2+36 s_1^2-8 s_2^2 s_3 s_1+16 s_2 s_3 s_1+40 s_3 s_1-16 s_3 s_4 s_1+2 s_2^4-12 s_2^2+8 s_2 s_3^2+16 s_3^2-16 s_2-8 s_2^2 s_4-16 s_2 s_4-8 s_4-6\right)-16384s_5^5 \left(s_1^3-2 s_2 s_1-4 s_1-2 s_3\right)+16384s_5^6=0$
其中$s_i$是$a_i$的初等对称多项式

kuing

回复 5# hbghlyj

没找到好方法。但是有恒等式：若$\arcsin a+\arcsin b+\arcsin c+\arcsin d=0\pmod{2\pi}$，则$s_1^4-4s_2s_1^2-4s_4s_1^2+8s_3s_1+4s_3^2=0$，其中$s_i$是a,b,c,d的初等对称多项式

那我 4# 就相当于推导了这个恒等式了。

在 4# 倒数第二条行间公式中，将 `(\sin a,\sin b,\sin c)` 分别写成 `(-y,-z,-w)`，即
\[(x^2-2xyzw-y^2-z^2+2y^2z^2+w^2)^2=4(1-y^2)(1-z^2)(xw-yz)^2,\]它正是等价于 `s_1^4-4s_2s_1^2-4s_4s_1^2+8s_3s_1+4s_3^2=0`，其中 `s_i` 是 `(x,y,z,w)` 也即 `(\sin(a+b+c),-\sin a,-\sin b,-\sin c)` 的初等对称多项式。

hejoseph

我的方法也是类似的，也是求出 $\sin\alpha$、$\sin\beta$、$\sin\gamma$、$\sin(\alpha+\beta+\gamma)$ 所满足的代数式。这种题目改一下数字就没法做了，太取巧。

由 $\sin(\alpha+\beta+\gamma)=\sin\alpha\cos\beta\cos\gamma+\sin\beta\cos\alpha\cos\gamma+\sin\gamma\cos\alpha\cos\gamma-\sin\alpha\sin\beta\sin\gamma$ 消去 $\cos\alpha$、$\cos\beta$、$\cos\gamma$ 得到一个代数式。为了方便，下面令 $t=\sin(\alpha+\beta+\gamma)$，$x=\sin\alpha$，$y=\sin\beta$，$z=\sin\gamma$，则得
\[
t^4+4xyzt^3-2(x^2+y^2+z^2-2x^2y^2-2x^2z^2-2y^2z^2)t^2+4xyz(x^2+y^2+z^2-2)t+x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2+4x^2y^2z^2=0
\]

ZCos666

5#中的第一条实际可以拓展：
\[ s_1^4-4s_2s_1^2+8s_3s_1+4s_3^2=\sin(a_1+a_2+a_3)\sin(a_1+a_2-a_3)\sin(a_2+a_3-a_1)\sin(a_3+a_1-a_2) \]
其中$s_i$是$\sin{a_i}$的初等对称多项式
另外还有，设$\theta=\dfrac{\alpha+\beta+\gamma}{2}$：
\[ \begin{aligned} \operatorname{C}_\alpha\operatorname{C}_\beta\operatorname{C}_\gamma&=\operatorname{C}_\theta\operatorname{C}_{\theta-\alpha}\operatorname{C}_{\theta-\beta}\operatorname{C}_{\theta-\gamma}+\operatorname{S}_\theta\operatorname{S}_{\theta-\alpha}\operatorname{S}_{\theta-\beta}\operatorname{S}_{\theta-\gamma}\\ \dfrac{1}{4}\left(\operatorname{C}_{2\alpha}+\operatorname{C}_{2\beta}+\operatorname{C}_{2\gamma}+1\right)&=\operatorname{C}_\theta\operatorname{C}_{\theta-\alpha}\operatorname{C}_{\theta-\beta}\operatorname{C}_{\theta-\gamma}-\operatorname{S}_\theta\operatorname{S}_{\theta-\alpha}\operatorname{S}_{\theta-\beta}\operatorname{S}_{\theta-\gamma} \end{aligned} \]
换个元有等价形式：
\[ \begin{aligned} \operatorname{C}_{2(\alpha+\beta)}\operatorname{C}_{2(\beta+\gamma)}\operatorname{C}_{2(\gamma+\alpha)}&=\operatorname{C}_{2\theta}\operatorname{C}_{\alpha}\operatorname{C}_{\beta}\operatorname{C}_{\gamma}+\operatorname{S}_{2\theta}\operatorname{S}_{\alpha}\operatorname{S}_{\beta}\operatorname{S}_{\gamma}\\ \dfrac{1}{4}\left(\operatorname{C}_{2(\alpha+\beta)}+\operatorname{C}_{2(\beta+\gamma)}+\operatorname{C}_{2(\gamma+\alpha)}+1\right)&=\operatorname{C}_{2\theta}\operatorname{C}_{\alpha}\operatorname{C}_{\beta}\operatorname{C}_{\gamma}-\operatorname{S}_{2\theta}\operatorname{S}_{\alpha}\operatorname{S}_{\beta}\operatorname{S}_{\gamma} \end{aligned} \]
至于后面两条，也就是四元及更多情形有没有同样优雅的恒等式，不太清楚