# Min max: y=sin x+cos x

• Karol

## Homework Equations

Minimum/Maximum occurs when the first derivative=0

## The Attempt at a Solution

$$y=\sin{x}+\cos{x}~\rightarrow~y'=\cos{x}-\sin{x}$$
$$y'=0:~\rightarrow~\cos{x}=\sin{x}~\rightarrow~x=\frac{\pi}{4}+n\cdot \pi$$
It's not correct

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It's not correct
Yes it is.

What makes you think it is not correct?

$$y(max)=45^0+2\pi k,~y(min)=225^0+2\pi k$$

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SammyS
$$y=\sin{x}+\cos{x}$$

There's a nice trig formula that gives:

##\sin x + \cos x = \sqrt{2}\sin(x + \frac{\pi}{4})##

Which hopefully settles the issue, if nothing else does.

Orodruin
There's a nice trig formula that gives:

##\sin x + \cos x = \sqrt{2}\sin(x + \frac{\pi}{4})##

Which hopefully settles the issue, if nothing else does.
Just to add some middle steps
$$\sin(x) + \cos(x) = \sqrt{2}[\sin(x)/\sqrt 2 + \cos(x)/\sqrt 2] = \sqrt 2 [\sin(x)\cos(\pi/4) + \cos(x)\sin(\pi/4)] = \sqrt 2 \sin(x+ \pi/4).$$

## y'=0:~\rightarrow~\cos{x}=\sin{x}~\rightarrow~x=\frac{\pi}{4}+n\cdot \pi ##
$$y(max)=45^0+2\pi k,~y(min)=225^0+2\pi k$$

Both agree. The original says, there is a max or min at ##\frac {\pi}{4} + n \pi \ \ \ ie. \frac {\pi}{4}, 5 \frac {\pi}{4}, 9 \frac {\pi}{4}, 13 \frac {\pi}{4}, etc. ##
The later simply separates them into max and min points with maxima at ## \frac {\pi}{4}, 9 \frac {\pi}{4}, 17 \frac {\pi}{4}, etc ##
and minima at ## 5 \frac {\pi}{4}, 13 \frac {\pi}{4}, etc. ##