高考模拟题

markfang2050

任何方法均可以。

zeroieme

\(\left\{v=\cos ^2B+\cos ^2C,\sin (B+C)-2 \cos B \cos C=0\right\}\)换元\(\left\{B\to \tan ^{-1}(x),C\to \tan ^{-1}(y)\right\}\)
得\(\left\{v=\frac{x^2+y^2+2}{\left(x^2+1\right) \left(y^2+1\right)},\frac{x+y-2}{\sqrt{x^2+1} \sqrt{y^2+1}}=0\right\}\)
对称有理式进一步换元\(\{x^2+y^2\to p,xy\to q\}\)
\(x+y=2\to p=4-2q\) 得\(v=\frac{p+2}{p+q^2+1}=\frac{6-2 q}{q^2-2 q+5}\)
求导后有\(q\to 3-2 \sqrt{2}\)时最大值为\(\frac{1}{2} \left(1+\sqrt{2}\right)\)

把\(q= 3-2 \sqrt{2}\)回代得， \(\left\{B=\tan ^{-1}\left(1-\sqrt{2 \left(\sqrt{2}-1\right)}\right)\approx 5.13 {}^{\circ},C=\tan ^{-1}\left(\sqrt{2 \left(\sqrt{2}-1\right)}+1\right)\approx62.37 {}^{\circ}\right\}\)或者\(\left\{B=\tan ^{-1}\left(\sqrt{2 \left(\sqrt{2}-1\right)}+1\right),C=\tan ^{-1}\left(1-\sqrt{2 \left(\sqrt{2}-1\right)}\right)\right\}\)

northwolves

由三角形中的恒等式          $cos^2A+cos^2B+cos^2C=1-2cosAcosBcosC$
可得$cos^2B+cos^2C=sin^2A-2cosAcosBcosC=sin^2A-sinAcosA=\frac{1}{2}-\frac{1}{2}cos2A-\frac{1}{2}sin2A=\frac{1}{2}-\frac{sqrt(2)}{2}sin(2A+π/4)<=1/2+\frac{sqrt(2)}{2}$,成立条件$A=\frac{5π}{8}$

markfang2050

解答：

northwolves

不记得或者不了解 3# 的恒等式的话，只好生推硬闯喽。
`\begin{split}\cos A&=-\cos(B+C)=\sin B\sin C-\cos B\cos C．\\
\cos^2A&=\sin^2B\sin^2C-2\sin B\sin C\cos B\cos C+\cos^2B\cos^2C\\
&=(1-\cos^2B)(1-\cos^2C)-2\sin B\sin C\cos B\cos C+\cos^2B\cos^2C\\
&=1-\cos^2B-\cos^2C+2\cos^2B\cos^2C-2\sin B\sin C\cos B\cos C\\
&=1-\cos^2B-\cos^2C+2\cos B\cos C(\cos B\cos C-\sin B\sin C)\\
&=1-\cos^2B-\cos^2C+2\cos B\cos C\cos(B+C)\\
&=1-\cos^2B-\cos^2C-\sin A\cos A\\

\cos^2B+\cos^2C&=1-\cos^2A-\sin A\cos A=\frac12-\frac12\cos2A-\frac12\sin2A\\&=\frac12-\frac{\sqrt2}{2}\sin(2A+π/4)\le\frac12+\frac{\sqrt2}{2}\end{split}`

northwolves

不记得或者不了解 3# 的恒等式的话，只好生推硬闯喽。
$cos^2B+cos^2C$
$=(1-cos^2B)(1-cos^2C)-cos^2Bcos^2C-1$
$=sin^2Bsin^2C-cos^2Bcos^2C-1$
$=(sinBsinC-cosBcosC)^2+2sinBsinCcosBcosC-2cos^2Bcos^2C-1$
$=cos^2(B+C)-2cosBcosC(cosBcosC-sinBsinC) -1$
$=cos^2A-sinAcos(B+C)-1$
$=sinAcosA-sin^2A$
$=(1-cos2A)/2-sin2A/2$
$=\frac{1}{2}-\frac{sqrt(2)}{2}sin(2A+π/4)$
$<=\frac{1}{2}+\frac{sqrt(2)}{2}$

mathematica

markfang2050 发表于 2019-4-23 22:33
解答：

1. Clear["Global`*"];
   
2. ans=Maximize[
   
3.     {Cos[B]^2+Cos[c]^2,
   
4.         Sin[A]==2*Cos[B]*Cos[c]&&
   
5.         0<A<Pi&&
   
6.         0<B<Pi&&
   
7.         0<c<Pi&&
   
8.         A+B+c==Pi
   
9.     },{A,B,c}
   
10. ]
    
11. aaa=FullSimplify[ans]
    

复制代码

计算结果
\[\left\{\cos ^2\left(2 \tan ^{-1}\left(\text{Root}\left[\left\{\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,\text{$\#$1}^2 \text{$\#$2}^4-2 \text{$\#$1}^2 \text{$\#$2}^2+\text{$\#$1}^2-\text{$\#$1} \text{$\#$2}^4-4 \text{$\#$1} \text{$\#$2}^3-2 \text{$\#$1} \text{$\#$2}^2+4 \text{$\#$1} \text{$\#$2}-\text{$\#$1}-\text{$\#$2}^4+2 \text{$\#$2}^2-1\&\right\},\{4,3\}\right]\right)+2 \tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,4\right]\right)\right)+\cos ^2\left(2 \tan ^{-1}\left(\text{Root}\left[\left\{\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,\text{$\#$1}^2 \text{$\#$2}^4-2 \text{$\#$1}^2 \text{$\#$2}^2+\text{$\#$1}^2-\text{$\#$1} \text{$\#$2}^4-4 \text{$\#$1} \text{$\#$2}^3-2 \text{$\#$1} \text{$\#$2}^2+4 \text{$\#$1} \text{$\#$2}-\text{$\#$1}-\text{$\#$2}^4+2 \text{$\#$2}^2-1\&\right\},\{4,3\}\right]\right)\right),\left\{A\to 2 \tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,4\right]\right),B\to 2 \tan ^{-1}\left(\text{Root}\left[\left\{\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,\text{$\#$1}^2 \text{$\#$2}^4-2 \text{$\#$1}^2 \text{$\#$2}^2+\text{$\#$1}^2-\text{$\#$1} \text{$\#$2}^4-4 \text{$\#$1} \text{$\#$2}^3-2 \text{$\#$1} \text{$\#$2}^2+4 \text{$\#$1} \text{$\#$2}-\text{$\#$1}-\text{$\#$2}^4+2 \text{$\#$2}^2-1\&\right\},\{4,3\}\right]\right),c\to -2 \tan ^{-1}\left(\text{Root}\left[\left\{\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,\text{$\#$1}^2 \text{$\#$2}^4-2 \text{$\#$1}^2 \text{$\#$2}^2+\text{$\#$1}^2-\text{$\#$1} \text{$\#$2}^4-4 \text{$\#$1} \text{$\#$2}^3-2 \text{$\#$1} \text{$\#$2}^2+4 \text{$\#$1} \text{$\#$2}-\text{$\#$1}-\text{$\#$2}^4+2 \text{$\#$2}^2-1\&\right\},\{4,3\}\right]\right)-2 \tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,4\right]\right)+\pi \right\}\right\}\]

化简结果
\[\left\{\sin ^2\left(2 \tan ^{-1}\left(\text{Root}\left[\left\{\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,\text{$\#$1}^2 \text{$\#$2}^4-2 \text{$\#$1}^2 \text{$\#$2}^2+\text{$\#$1}^2-\text{$\#$1} \text{$\#$2}^4-4 \text{$\#$1} \text{$\#$2}^3-2 \text{$\#$1} \text{$\#$2}^2+4 \text{$\#$1} \text{$\#$2}-\text{$\#$1}-\text{$\#$2}^4+2 \text{$\#$2}^2-1\&\right\},\{4,3\}\right]\right)+\frac{\pi }{8}\right)+\cos ^2\left(2 \tan ^{-1}\left(\text{Root}\left[\left\{\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,\text{$\#$1}^2 \text{$\#$2}^4-2 \text{$\#$1}^2 \text{$\#$2}^2+\text{$\#$1}^2-\text{$\#$1} \text{$\#$2}^4-4 \text{$\#$1} \text{$\#$2}^3-2 \text{$\#$1} \text{$\#$2}^2+4 \text{$\#$1} \text{$\#$2}-\text{$\#$1}-\text{$\#$2}^4+2 \text{$\#$2}^2-1\&\right\},\{4,3\}\right]\right)\right),\left\{A\to \frac{5 \pi }{8},B\to 2 \tan ^{-1}\left(\text{Root}\left[\left\{\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,\text{$\#$1}^2 \text{$\#$2}^4-2 \text{$\#$1}^2 \text{$\#$2}^2+\text{$\#$1}^2-\text{$\#$1} \text{$\#$2}^4-4 \text{$\#$1} \text{$\#$2}^3-2 \text{$\#$1} \text{$\#$2}^2+4 \text{$\#$1} \text{$\#$2}-\text{$\#$1}-\text{$\#$2}^4+2 \text{$\#$2}^2-1\&\right\},\{4,3\}\right]\right),c\to \frac{3 \pi }{8}-2 \tan ^{-1}\left(\text{Root}\left[\left\{\text{$\#$1}^4+4 \text{$\#$1}^3-6 \text{$\#$1}^2-4 \text{$\#$1}+1\&,\text{$\#$1}^2 \text{$\#$2}^4-2 \text{$\#$1}^2 \text{$\#$2}^2+\text{$\#$1}^2-\text{$\#$1} \text{$\#$2}^4-4 \text{$\#$1} \text{$\#$2}^3-2 \text{$\#$1} \text{$\#$2}^2+4 \text{$\#$1} \text{$\#$2}-\text{$\#$1}-\text{$\#$2}^4+2 \text{$\#$2}^2-1\&\right\},\{4,3\}\right]\right)\right\}\right\}\]

数值解的结果
{1.2071067811865475244008443621048490392848359376884740365883398689953\
66239231053519425193767163820786,
{A ->
   1.96349540849362077403915211454968930262323087460944113810934037019\
2385253928880624142521765838823167,
  B -> 0.0895798942512430460337534906252297154364890497406960087899781\
9792170824985017997107414335071478519798,
  c -> 1.0885173508449294183897377781045838661374494750249686740756260\
24193722902507148403411369708788508703}}

mathematica

mathematica 发表于 2020-4-21 16:07
计算结果
\[\left\{\cos ^2\left(2 \tan ^{-1}\left(\text{Root}\left[\left\{\text{$\#$1}^4+4 \te ...

1. Clear["Global`*"];
   
2. ans=Maximize[
   
3.     {1/(1+x^2)+1/(1+y^2),
   
4.         x+y==2
   
5.     },{x,y}
   
6. ]//FullSimplify//ToRadicals
   

复制代码

求解结果：

\[\left\{\frac{1}{2}+\frac{1}{\sqrt{2}},\left\{x\to \sqrt{2 \left(\sqrt{2}-1\right)}+1,y\to 1-\sqrt{2 \left(\sqrt{2}-1\right)}\right\}\right\}\]

hujunhua

@zeroieme
在2#得到`\D v=\frac{6-2q}{q^2-2q+5}`后，可以化成   \(vq^2-2(v-1)q+(5v-6)=0\)

然后由判别式   `\D\Delta=4(v-1)^2-4v(5v-6)\ge0\to v\le\frac{\sqrt2+1}2`.

@markfang2050
记 `v=\cos^2B+\cos^2C`, 4#的结果可以进一步化为 \(\D v=\frac{2x^2+4}{x^4+4}\\vx^4-2x^2+4(v-1)=0\)

然后由判别式   `\D\Delta=4-16v(v-1)\ge0\to v\le\frac{\sqrt2+1}2`.
中学生一般不会错过这种方法。

hujunhua

northwolves 发表于 2019-4-23 23:01
不记得或者不了解 3# 的恒等式的话，只好生推硬闯喽。
`\begin{split}\cos A&=-\cos(B+C)=\sin B\sin C-\c ...

不了解 3# 的那个恒等式的话，也不要这样一头扎进去硬闯，可以作代换
`A'=\pi-A, B'=\pi/2-B, C'=\pi/2-C`, 将题目转化为
已知`\sin A'=2\sin B'\sin C'`，求 `\sin^2B'+\sin^2C'`的最大值。
由余弦定理得 `\sin^2B'+\sin^2C'=\sin^2A'+2\sin B'\sin C'\cos A'=\sin^2A'+\sin A'\cos A'`.