Finding Minimum and Maximum with Trigonometric Functions: y=sin x+cos x

[Karol]

Homework Statement

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

 

[Orodruin]

It's not correct

Yes it is.

 

[fresh_42]

What makes you think it is not correct?

 

[Karol]

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

 

[Orodruin]

$$y(max)=45^0+2\pi k,~y(min)=225^0+2\pi k$$
View attachment 231062

$\pi/4 \simeq 0.785$

 

[PeroK]

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

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
$$
\sin(x) + \cos(x) = \sqrt{2}[\sin(x)/\sqrt 2 + \cos(x)/\sqrt 2] = \sqrt 2 [\sin(x)\cos(\pi/4) + \cos(x)\sin(\pi/4)] = \sqrt 2 \sin(x+ \pi/4).
$$

 

[Merlin3189]

$y'=0:~\rightarrow~\cos{x}=\sin{x}~\rightarrow~x=\frac{\pi}{4}+n\cdot \pi$

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

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 $5 \frac {\pi}{4}, 13 \frac {\pi}{4}, etc.$

Your first answer was not wrong, just inadequate if you were required to list them separately.