Question

The minimum value of $\sin \left(x\right)+\cos \left(x\right)$ is

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $-\frac{1}{\sqrt{2}}$
• E. $1$

Answer 1

The correct option is B

$-\sqrt{2}$

Find the minimum value of the given function

Given : $f\left(x\right)=\sin \left(x\right)+\cos \left(x\right)$

Multiply and divide by $\sqrt{2}$,

$\begin{array}{rcl}f\left(x\right)& =& \sqrt{2}\left[\frac{1}{\sqrt{2}}\sin \left(x\right)+\frac{1}{\sqrt{2}}\cos \left(x\right)\right]\left[\sin \left(45°\right)=\cos \left(45°\right)=\frac{1}{\sqrt{2}}\right]\\ & =& \sqrt{2}\left[\sin \left(x\right)\cos \left(45°\right)+\cos \left(x\right)\sin \left(45°\right)\right]\left[\sin \left(a+b\right)=\sin \left(a\right)\cos \left(b\right)+\cos \left(a\right)\sin \left(b\right)\right]\\ & =& \sqrt{2}\left[\sin \left(x+45°\right)\right]\end{array}$

Since the minimum value of $\sin \left(x\right)$ is $-1$.

Therefore, $\begin{array}{rcl}f{\left(x\right)}_{min}& =& \sqrt{2}\left[-1\right]\end{array}$

$\Rightarrow$ $f{\left(x\right)}_{min}=-\sqrt{2}$

Hence option $\left(B\right)$ is the correct answer.