题目介绍

本题给出一个三角函数的等式，要求证明另一个三角函数的表达式与之相等。需要运用三角函数的基本性质以及代数计算的技巧，最终得出结论。

解题思路

首先根据三角函数的定义展开等式：

$\begin{aligned} \text{左边} & = \sin \theta + \cos \theta \ \text{右边} & = \pm \sqrt 2 \cos A \ \text{原式} & = \tan A \ & = \frac{\sin A}{\cos A} \ & = \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} \cdot \frac{\sin \theta + \cos \theta}{\sin \theta + \cos \theta} \ & = \frac{(\sin \theta + \cos \theta)^2}{\sin^2 \theta - \cos^2 \theta} \end{aligned}$

接下来可以运用三角函数的基本公式变形：

$\sin^2 \theta + \cos^2 \theta = 1, \sin^2 \theta = 1 - \cos^2 \theta$

将其代入原式：

$\begin{aligned} \text{原式} & = \frac{(\sin \theta + \cos \theta)^2}{\sin^2 \theta - \cos^2 \theta} \ & = \frac{(\sin \theta + \cos \theta)^2}{1 - 2 \cos^2 \theta} \end{aligned}$

根据等式 $\sin A = \sin (180 - A) = \cos A$：

$\begin{aligned} \text{原式} & = \frac{(\sin \theta + \cos \theta)^2}{1 - 2 \sin^2 (\frac{\pi}{2} - \theta)} \ & = \frac{(\sin \theta + \cos \theta)^2}{1 - 2 \sin^2 (\frac{\pi}{4} + \frac{\theta}{2})} \end{aligned}$

将等式 $1 - 2t^2 = (1 - \sqrt 2 t)(1 + \sqrt 2 t)$ 代入上式：

$\begin{aligned} \text{原式} & = \frac{(\sin \theta + \cos \theta)^2}{(1 - \sqrt 2 \sin (\frac{\pi}{4} + \frac{\theta}{2}))(1 + \sqrt 2 \sin (\frac{\pi}{4} + \frac{\theta}{2}))} \ & = \frac{(\sin \theta + \cos \theta)^2}{(\cos (\frac{\pi}{4} - \frac{\theta}{2}) - \sin (\frac{\pi}{4} - \frac{\theta}{2}))(\cos (\frac{\pi}{4} + \frac{\theta}{2}) - \sin (\frac{\pi}{4} + \frac{\theta}{2}))} \end{aligned}$

继续变形：

$\begin{aligned} \text{原式} & = \frac{(\sin \theta + \cos \theta)^2}{\cos \frac{\pi}{4} - \sin \frac{\pi}{4} - (\cos \frac{\pi}{4} - \sin \frac{\pi}{4}) \cos \theta + (\cos \frac{\pi}{4} + \sin \frac{\pi}{4})\sin \theta \cdot (\cos \frac{\pi}{4} + \sin \frac{\pi}{4})\cos \theta - (\cos \frac{\pi}{4} + \sin \frac{\pi}{4})\sin \theta} \ & = \frac{(\sin \theta + \cos \theta)^2}{-\sqrt 2 \cos \theta + (\sqrt 2 - 1)\sin \theta} \ & = \frac{\sqrt 2 \cos \theta - (\sqrt 2 - 1)\sin \theta}{\sqrt 2 \cos \theta - (\sqrt 2 - 1)\sin \theta} \cdot \frac{(\sin \theta + \cos \theta)^2}{-\sqrt 2 \cos \theta + (\sqrt 2 - 1)\sin \theta} \ & = \frac{(1 - \sqrt 2 \tan \theta)(\sin \theta + \cos \theta)^2}{-1 + (\sqrt 2 + 1)\tan \theta} \ & = \frac{\frac{\sin \theta + \cos \theta}{\cos \theta}}{-\frac{1}{\cos \theta} + \sqrt 2 + \frac{1}{\cos \theta}} \cdot (\sin \theta + \cos \theta)^2 \ & = \frac{\sin \theta + \cos \theta}{\sqrt 2 \cos \theta - \sin \theta} \cdot (\sin \theta + \cos \theta)^2 \ & = \pm \sqrt 2 \cos A \end{aligned}$

根据 $\sin A = \cos (\frac{\pi}{2} - A)$ 可以得出：

$\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt 2}$

因此：

$\begin{aligned} \text{左边} & = \sqrt 2 \ \text{右边} & = \pm \sqrt 2 \cos A \ & = \pm \sqrt 2 \sin \frac{\pi}{4} \ & = \pm \sqrt 2 \cdot \frac{1}{\sqrt 2} \ & = \pm 1 \end{aligned}$

从而得出：

$\sin \theta + \cos \theta = \pm \sqrt 2 \cos A$

结论

根据上述推导过程，可以得到 $\sin \theta + \cos \theta = \pm \sqrt 2 \cos A$。其中，正负号由右边的 $\pm$ 决定。