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一道三角函数题

题目:已知$\alpha,\beta$是锐角,且$\sin(\alpha+\beta)=\sin^2\alpha+\sin^2\beta$.求证:$\alpha+\beta=\frac{\pi}{2}$.

:
\begin{align*}
\sin(\alpha+\beta)=\sin^2\alpha+\sin^2\beta&\iff
\sin(\alpha+\beta)=\frac{1-\cos2\alpha}{2}+\frac{1-\cos2\beta}{2}=1-\frac{1}{2}(\cos2\alpha+\cos2\beta)
\\&\iff \sin(\alpha+\beta)=1-\cos(\alpha+\beta)\cos(\alpha-\beta)
\end{align*}
假如$\alpha+\beta\neq \frac{\pi}{2}$,则可得
$$
\cos(\alpha-\beta)=\frac{1-\sin(\alpha+\beta)}{\cos(\alpha+\beta)}=\frac{1-\cos
\left[\frac{\pi}{2}-(\alpha+\beta)\right]}{\sin
[\frac{\pi}{2}-(\alpha+\beta)]}=\frac{2\sin^2(\frac{\pi}{4}-\frac{\alpha+\beta}{2})}{2\sin
(\frac{\pi}{4}-\frac{\alpha+\beta}{2})\cos
(\frac{\pi}{4}-\frac{\alpha+\beta}{2})}=\tan\left(\frac{\pi}{4}-\frac{\alpha+\beta}{2}\right).
$$
令$\alpha-\beta=x_{1},\alpha+\beta=x_2$,则
\begin{equation}\label{eq:1}
\cos x_1=\tan\left(\frac{\pi}{4}-\frac{x_2}{2}\right),
\end{equation}
其中$x_1\in (-\frac{\pi}{2},\frac{\pi}{2})$,$x_2\in (0,\pi)$.进一步的,由式\eqref{eq:1},可得
$$
\tan\left(\frac{\pi}{4}-\frac{x_2}{2}\right)>0\Rightarrow 0 < x_2 < \frac{\pi}{2}.
$$
因此有
$$
-\frac{\pi}{2} < x_1 < x_2 < \frac{\pi}{2}.
$$
结合函数$y=\tan(\frac{\pi}{4}-\frac{x}{2})$与函数$y=\cos x$在区间$(-\frac{\pi}{2},\frac{\pi}{2})$的图像1,可得,在区间$(-\frac{\pi}{2},\frac{\pi}{2})$内,要让式\eqref{eq:1}成立,必须要有$x_2 < x_1$,这与$x_1 < x_2$矛盾.

可见,假设不成立,因此只能有$\alpha+\beta=\frac{\pi}{2}$.且当$\alpha+\beta=\frac{\pi}{2}$时,确实有$\sin(\alpha+\beta)=\sin^2\alpha+\sin^2\beta$.
图1