如何求下面的函数的极值？

mathematica

求下面这个表达式的极值\[
y=\frac{\sin (a+x) \sin (b+x)}{\sin (c+x) \sin (d+x)}
\]令导数=0易得驻点条件\[
\frac{\sin (a+x) \sin (b+x)}{\sin (c+x) \sin (d+x)}=\frac{\sin(2x+a+b)}{\sin(2x+c+d)}
\]怎么解这个驻点方程呢？

题目背景：土木专业，求岩土边坡上的破裂角

northwolves

K>> syms a b c d x y
y=sin(x+a)*sin(x+b)/(sin(x+c)*sin(x+d))
solve(diff(y))

有解的情况下，返回
$
ans =
     -log(-((exp(a*2i)*exp(b*2i) - exp(c*2i)*exp(d*2i) + ((exp(a*2i) - exp(c*2i))*(exp(a*2i) - exp(d*2i))*(exp(b*2i) - exp(c*2i))*(exp(b*2i) - exp(d*2i)))^(1/2))/(exp(a*2i)*exp(b*2i)*exp(c*2i) + exp(a*2i)*exp(b*2i)*exp(d*2i) - exp(a*2i)*exp(c*2i)*exp(d*2i) - exp(b*2i)*exp(c*2i)*exp(d*2i)))^(1/2))*1i
-log(-(-(- exp(a*2i)*exp(b*2i) + exp(c*2i)*exp(d*2i) + ((exp(a*2i) - exp(c*2i))*(exp(a*2i) - exp(d*2i))*(exp(b*2i) - exp(c*2i))*(exp(b*2i) - exp(d*2i)))^(1/2))/(exp(a*2i)*exp(b*2i)*exp(c*2i) + exp(a*2i)*exp(b*2i)*exp(d*2i) - exp(a*2i)*exp(c*2i)*exp(d*2i) - exp(b*2i)*exp(c*2i)*exp(d*2i)))^(1/2))*1i
     -log(((exp(a*2i)*exp(b*2i) - exp(c*2i)*exp(d*2i) + ((exp(a*2i) - exp(c*2i))*(exp(a*2i) - exp(d*2i))*(exp(b*2i) - exp(c*2i))*(exp(b*2i) - exp(d*2i)))^(1/2))/(exp(a*2i)*exp(b*2i)*exp(c*2i) + exp(a*2i)*exp(b*2i)*exp(d*2i) - exp(a*2i)*exp(c*2i)*exp(d*2i) - exp(b*2i)*exp(c*2i)*exp(d*2i)))^(1/2))*1i
  -log((-(- exp(a*2i)*exp(b*2i) + exp(c*2i)*exp(d*2i) + ((exp(a*2i) - exp(c*2i))*(exp(a*2i) - exp(d*2i))*(exp(b*2i) - exp(c*2i))*(exp(b*2i) - exp(d*2i)))^(1/2))/(exp(a*2i)*exp(b*2i)*exp(c*2i) + exp(a*2i)*exp(b*2i)*exp(d*2i) - exp(a*2i)*exp(c*2i)*exp(d*2i) - exp(b*2i)*exp(c*2i)*exp(d*2i)))^(1/2))*1i$

chyanog

\(x\to \arctan \left(-\tan (a)-\tan (b)+\tan (c)+\tan (d),\ \tan (a) \tan (b)-\tan (c) \tan (d)+\sqrt{(\tan (a)-\tan (c)) (\tan (b)-\tan (c)) (\tan (a)-\tan (d)) (\tan (b)-\tan (d))}\right)\)
\(x\to \arctan \left(\tan (a)+\tan (b)-\tan (c)-\tan (d),\ -\tan (a) \tan (b)+\tan (c) \tan (d)+\sqrt{(\tan (a)-\tan (c)) (\tan (b)-\tan (c)) (\tan (a)-\tan (d)) (\tan (b)-\tan (d))}\right)\)
Mathematica检验代码

1. Sin[c+x] Sin[d+x] Sin[a+b+2 x]-Sin[a+x] Sin[b+x] Sin[c+d+2 x] /.{
   
2.    {x->ArcTan[-Tan[a]-Tan[b]+Tan[c]+Tan[d], Tan[a] Tan[b]-Tan[c] Tan[d]+Sqrt[(Tan[a]-Tan[c]) (Tan[b]-Tan[c]) (Tan[a]-Tan[d]) (Tan[b]-Tan[d])]]},
   
3.    {x->ArcTan[Tan[a]+Tan[b]-Tan[c]-Tan[d], -Tan[a] Tan[b]+Tan[c] Tan[d]+Sqrt[(Tan[a]-Tan[c]) (Tan[b]-Tan[c]) (Tan[a]-Tan[d]) (Tan[b]-Tan[d])]]}
   
4.   } // TrigExpand // Simplify

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输出

{0, 0}

mathematica

chyanog 发表于 2019-11-11 10:08
\(x\to \arctan \left(-\tan (a)-\tan (b)+\tan (c)+\tan (d), \tan (a) \tan (b)-\tan (c) \tan (d)+\sqrt ...

牛人，你这个怎么算出来的，
你这个ArcTan[-2,-1]是180度+atan0.5吗？

mathematica

chyanog 发表于 2019-11-11 10:08
\(x\to \arctan \left(-\tan (a)-\tan (b)+\tan (c)+\tan (d),\ \tan (a) \tan (b)-\tan (c) \tan (d)+\sqr ...

1. Sin[c+x] Sin[d+x] Sin[a+b+2 x]-Sin[a+x] Sin[b+x] Sin[c+d+2 x]/.{{x->ArcTan[-Tan[a]-Tan[b]+Tan[c]+Tan[d],Tan[a] Tan[b]-Tan[c] Tan[d]-Sqrt[(Tan[a]-Tan[c]) (Tan[b]-Tan[c]) (Tan[a]-Tan[d]) (Tan[b]-Tan[d])]]},
   
2. {x->ArcTan[Tan[a]+Tan[b]-Tan[c]-Tan[d],-Tan[a] Tan[b]+Tan[c] Tan[d]-Sqrt[(Tan[a]-Tan[c]) (Tan[b]-Tan[c]) (Tan[a]-Tan[d]) (Tan[b]-Tan[d])]]}}//TrigExpand//Simplify

复制代码

如我所猜，这两个也是解！

mathematica

mathematica 发表于 2019-11-11 11:07
如我所猜，这两个也是解！

1. Clear["Global`*"];(*Clear all variables*)
   
2. y1=Sin[x+1]*Sin[x+2]
   
3. y2=Sin[x+3]*Sin[x+4]
   
4. yx=D[y1,x]*y2-y1*D[y2,x]
   
5. Reduce[yx==0,{x}]//FullSimplify
   

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输出结果
\[c_1\in \mathbb{Z}\land \left(x=2 \left(\tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4 (\sin (1) \sin (2) \sin (3) \cos (4)-\sin (1) \sin (4) \sin (3) \cos (2)-\sin (2) \sin (4) \sin (3) \cos (1)+\sin (1) \sin (2) \sin (4) \cos (3))+\text{$\#$1}^3 (4 \sin (3) \sin (4) \cos (1) \cos (2)-4 \sin (1) \sin (2) \cos (3) \cos (4))+\text{$\#$1}^2 (4 \sin (1) \cos (2) \cos (3) \cos (4)-2 \sin (2) \sin (3) \sin (1) \cos (4)-2 \sin (2) \sin (4) \sin (1) \cos (3)+2 \sin (3) \sin (4) \sin (1) \cos (2)+4 \sin (2) \cos (1) \cos (3) \cos (4)-4 \sin (3) \cos (1) \cos (2) \cos (4)-4 \sin (4) \cos (1) \cos (2) \cos (3)+2 \sin (2) \sin (3) \sin (4) \cos (1))+\text{$\#$1} (4 \sin (1) \sin (2) \cos (3) \cos (4)-4 \sin (3) \sin (4) \cos (1) \cos (2))+\sin (1) \sin (2) \sin (3) \cos (4)+\sin (1) \sin (2) \sin (4) \cos (3)-\sin (1) \sin (3) \sin (4) \cos (2)-\sin (2) \sin (3) \sin (4) \cos (1)\&,1\right]\right)+\pi  c_1\right)\lor x=2 \left(\tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4 (\sin (1) \sin (2) \sin (3) \cos (4)-\sin (1) \sin (4) \sin (3) \cos (2)-\sin (2) \sin (4) \sin (3) \cos (1)+\sin (1) \sin (2) \sin (4) \cos (3))+\text{$\#$1}^3 (4 \sin (3) \sin (4) \cos (1) \cos (2)-4 \sin (1) \sin (2) \cos (3) \cos (4))+\text{$\#$1}^2 (4 \sin (1) \cos (2) \cos (3) \cos (4)-2 \sin (2) \sin (3) \sin (1) \cos (4)-2 \sin (2) \sin (4) \sin (1) \cos (3)+2 \sin (3) \sin (4) \sin (1) \cos (2)+4 \sin (2) \cos (1) \cos (3) \cos (4)-4 \sin (3) \cos (1) \cos (2) \cos (4)-4 \sin (4) \cos (1) \cos (2) \cos (3)+2 \sin (2) \sin (3) \sin (4) \cos (1))+\text{$\#$1} (4 \sin (1) \sin (2) \cos (3) \cos (4)-4 \sin (3) \sin (4) \cos (1) \cos (2))+\sin (1) \sin (2) \sin (3) \cos (4)+\sin (1) \sin (2) \sin (4) \cos (3)-\sin (1) \sin (3) \sin (4) \cos (2)-\sin (2) \sin (3) \sin (4) \cos (1)\&,2\right]\right)+\pi  c_1\right)\lor x=2 \left(\tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4 (\sin (1) \sin (2) \sin (3) \cos (4)-\sin (1) \sin (4) \sin (3) \cos (2)-\sin (2) \sin (4) \sin (3) \cos (1)+\sin (1) \sin (2) \sin (4) \cos (3))+\text{$\#$1}^3 (4 \sin (3) \sin (4) \cos (1) \cos (2)-4 \sin (1) \sin (2) \cos (3) \cos (4))+\text{$\#$1}^2 (4 \sin (1) \cos (2) \cos (3) \cos (4)-2 \sin (2) \sin (3) \sin (1) \cos (4)-2 \sin (2) \sin (4) \sin (1) \cos (3)+2 \sin (3) \sin (4) \sin (1) \cos (2)+4 \sin (2) \cos (1) \cos (3) \cos (4)-4 \sin (3) \cos (1) \cos (2) \cos (4)-4 \sin (4) \cos (1) \cos (2) \cos (3)+2 \sin (2) \sin (3) \sin (4) \cos (1))+\text{$\#$1} (4 \sin (1) \sin (2) \cos (3) \cos (4)-4 \sin (3) \sin (4) \cos (1) \cos (2))+\sin (1) \sin (2) \sin (3) \cos (4)+\sin (1) \sin (2) \sin (4) \cos (3)-\sin (1) \sin (3) \sin (4) \cos (2)-\sin (2) \sin (3) \sin (4) \cos (1)\&,3\right]\right)+\pi  c_1\right)\lor x=2 \left(\tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4 (\sin (1) \sin (2) \sin (3) \cos (4)-\sin (1) \sin (4) \sin (3) \cos (2)-\sin (2) \sin (4) \sin (3) \cos (1)+\sin (1) \sin (2) \sin (4) \cos (3))+\text{$\#$1}^3 (4 \sin (3) \sin (4) \cos (1) \cos (2)-4 \sin (1) \sin (2) \cos (3) \cos (4))+\text{$\#$1}^2 (4 \sin (1) \cos (2) \cos (3) \cos (4)-2 \sin (2) \sin (3) \sin (1) \cos (4)-2 \sin (2) \sin (4) \sin (1) \cos (3)+2 \sin (3) \sin (4) \sin (1) \cos (2)+4 \sin (2) \cos (1) \cos (3) \cos (4)-4 \sin (3) \cos (1) \cos (2) \cos (4)-4 \sin (4) \cos (1) \cos (2) \cos (3)+2 \sin (2) \sin (3) \sin (4) \cos (1))+\text{$\#$1} (4 \sin (1) \sin (2) \cos (3) \cos (4)-4 \sin (3) \sin (4) \cos (1) \cos (2))+\sin (1) \sin (2) \sin (3) \cos (4)+\sin (1) \sin (2) \sin (4) \cos (3)-\sin (1) \sin (3) \sin (4) \cos (2)-\sin (2) \sin (3) \sin (4) \cos (1)\&,4\right]\right)+\pi  c_1\right)\right)\]

markfang2050

chyanog 发表于 2019-11-11 10:08
\(x\to \arctan \left(-\tan (a)-\tan (b)+\tan (c)+\tan (d),\ \tan (a) \tan (b)-\tan (c) \tan (d)+\sqr ...

这个解怎么算出来的，代码？

chyanog

markfang2050 发表于 2019-11-11 13:45
这个解怎么算出来的，代码？

1. Solve[D[(Sin[x + a] Sin[x + b])/(Sin[x + c] Sin[x + d]), x] == 0 /. x -> ArcTan[t] // TrigExpand, t]
   
2. % /. Sin[x_] :> Cos[x] HoldForm@Tan[x] // Simplify
   
3. % /. k_ Sqrt[x_] :> Sqrt[k^2 x] // Simplify // ReleaseHold

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markfang2050

chyanog 发表于 2019-11-11 14:00

学习了。

mathematica

chyanog 发表于 2019-11-11 14:00

x -> ArcTan[t]
你这么一替换，就漏解了，
因为ArcTan[t]的范围是从-90度到90度，
而x的范围没这个限制