\(x=125/4\) \(x=125/3\)¡

ͼ
\[
\left.\begin{array}{l}
\displaystyle
\cos\theta=\frac{25}{x}=\frac{12}{a}
\\
\displaystyle
\sin\theta=\frac{a}{25}
\end{array}\right\}\Rightarrow~\sin 2\theta=2\cdot\frac{a}{25}\cdot\frac{12}{a}=\frac{24}{25}~\Rightarrow~\cos2\theta=\pm\frac{7}{25}
\]
е \(\pm\) ȡ \(2\theta\) Ϊǻ۽ǡҺİǹʽ
\[
\cos\theta=\sqrt{\frac{1+\cos 2\theta}{2}}=\left\{
\begin{array}{ll}
\displaystyle
\frac{4}{5}, & 0<\theta<\pi/4
\\
\displaystyle
\frac{3}{5}, & \pi/4<\theta<\pi/2
\end{array}\right.
\]
ǵ
\[
x=\frac{25}{\cos\theta}=\left\{\begin{array}{ll}
\displaystyle
\frac{125}{4}, & 0<\theta<\pi/4
\\
\displaystyle
\frac{125}{3}, & \pi/4<\theta<\pi/2
\end{array}\right.
\]
ͼ
\[
\left.\begin{array}{l}
\displaystyle
\cos\theta=\frac{25}{x}=\frac{12}{a}
\\
\displaystyle
\sin\theta=\frac{a}{25}
\end{array}\right\}\Rightarrow~\sin 2\theta=2\cdot\frac{a}{25}\cdot\frac{12}{a}=\frac{24}{25}~\Rightarrow~\cos2\theta=\pm\frac{7}{25}
\]
е \(\pm\) ȡ \(2\theta\) Ϊǻ۽ǡҺİǹʽ
\[
\cos\theta=\sqrt{\frac{1+\cos 2\theta}{2}}=\left\{
\begin{array}{ll}
\displaystyle
\frac{4}{5}, & 0<\theta<\pi/4
\\
\displaystyle
\frac{3}{5}, & \pi/4<\theta<\pi/2
\end{array}\right.
\]
ǵ
\[
x=\frac{25}{\cos\theta}=\left\{\begin{array}{ll}
\displaystyle
\frac{125}{4}, & 0<\theta<\pi/4
\\
\displaystyle
\frac{125}{3}, & \pi/4<\theta<\pi/2
\end{array}\right.
\]
���༭ʱ��: 2021-11-07 07:17:51

