Maximum value in trigonometry

Let $f\left( x \right)=\sin \left( x \right)+\cos \left( x \right)$ , $x\in \mathbb{R}$ .

Then,
$\begin{gathered} & f\left( x \right) = \sqrt 2 \cdot \left( {\frac{{\sqrt 2 }}{2}\sin \left( x \right) + \frac{{\sqrt 2 }}{2}\cos \left( x \right)} \right) \\ & = \sqrt 2 \cdot \left( {\sin \left( x \right)\cos \left( {\frac{\pi }{4}} \right) + \sin \left( {\frac{\pi }{4}} \right)\cos \left( x \right)} \right) \\ & = \sqrt 2 \cdot \sin \left( {x + \frac{\pi }{4}} \right) \\ & \leqslant \sqrt 2 \\ \end{gathered}$

$\begin{gathered} & f\left( \frac{\pi }{4} \right)=\sin \left( \frac{\pi }{4} \right)+\cos \left( \frac{\pi }{4} \right) \\ & =\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} \\ & =\sqrt{2}\simeq \text{1}\text{.4142} \\ \end{gathered}$

Hence, $f\left( x \right) \leqslant f\left( {\frac{\pi }{4}} \right)$ , for all $x \in \mathbb{R}$ , thus, the maximum value of $\sin{(x)} + \cos{(x)}$ to three significant figures is $\boxed{1.41}$ .

$-\sqrt 2 \leq \sin x+\cos x=\sqrt 2 \sin (x+45\degree)\leq \sqrt 2$ .

So, the maximum value of $\sin x+\cos x$ is $\sqrt 2 \approx \boxed {1.414}$ .