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# What is the maximum value of the function $\sin \: x + \cos x$?

$\begin{array}{1 1} \text{maximum value of}f(x)=\sqrt 2 \\ \text{maximum value of} f(x)= 0 \\\text{minimum value of} f(x)=\sqrt 2 \\ \text{maximum value of }f(x)= -\sqrt 2 \end{array}$

Toolbox:
• $\large\frac{d}{dx}$$(\cos x)=-\sin x • \large\frac{d}{dx}$$(\sin x)=\cos x$
Step 1:
Let $f(x)=\sin x+\cos x$
On differentiating we get
$f'(x)=\cos x-\sin x$
For maxima and minima
$f'(x)=0$
$\Rightarrow \cos x-\sin x=0$
$\cos x[1-\tan x]=0$
$\tan x=1$
$x=\tan^{-1}1$
$x=\large\frac{\pi}{4}$ or $\large\frac{5\pi}{4}$
Step 2:
Now we find $f(x)$ at $x=0,\large\frac{\pi}{4},\frac{5\pi}{4}$$,2\pi f(0)=\sin 0+\cos 0=1 f(\large\frac{\pi}{4})$$=\sin \large\frac{\pi}{4}+$$\cos\large\frac{\pi}{4}$$=\sqrt 2$
$f(\large\frac{5\pi}{4})$$=\sin \large\frac{5\pi}{4}+$$\cos\large\frac{5\pi}{4}$
$\qquad=\large\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}$
$\qquad=\large\frac{-2}{\sqrt 2}$
$\qquad=-\sqrt 2$
$f(2\pi)=\sin 2\pi+\cos 2\pi$
$\qquad\;=1$
Hence the maximum value of $f(x)=\sqrt 2$

edited Aug 30, 2013