min and max of trig function I'm trying to help a friend of mine solve this problem. It's been like 10 years since I took calculus please help:
Find min/max values of $y = \sin x + \cos x$ on  $[0, 2\pi)$
Thanks.
I've got this much:


*

*$f'(x) = \cos x - \sin x$. 

*$0 = \cos x - \sin x$. 

*$\sin x = \cos x$. 

*$\tan x = 1$.


Thanks
Evaulate at end points and at the critical point:


*

*$f(0) = 1$

*$f(2\pi) = 1$

*$\sin x = \cos x$ at $\pi/4$


Hence,
$f(x)$ has a minimum at $\pi/4$ and $f(x)$ has a maximum at $x = 0$ on $[0,2\pi)$ (Extreme Value Theorem) 
 A: No, you don't get $\tan x = 0$; you would get, at best, $\tan x = 1$ (remember that $\tan x = \frac{\sin x}{\cos x}$. If $\sin x$ and $\cos x$ are equal, then the quotient is equal to $1$, not to $0$).
So, you remembered that you want to find the critical points (points where the derivative is zero). That's good. What you also need to remember is that if you have a continuous function on a closed interval, then the maximum and the minimum will each be achieved at either a critical point or an endpoint.
Here, you might as well work over $[0,2\pi]$ (the value at $2\pi$ is the same as the value at $0$). So the maximum and the minimum of $f(x)$ will be achieved either at $x=0$, or at a point where $f'(x)=0$, that is, a point where $\sin(x) = \cos(x)$ in the interval (better to work with these, since this equality does not depend on $\cos(x)\neq 0$, whereas $\tan(x)=1$ does; of course, it does not really matter here because if $\cos(x)=0$, then $\sin(x)\neq \cos(x)$). 
So, the question is: for what points $x$, $0\leq x\leq 2\pi$, do you have $\sin(x)=\cos(x)$? There are two such points; once you have them, simply evaluate the original function at these points, and at $x=0$ (where it has the same value as at $x=2\pi$, which is why we were able to add $2\pi$ to the interval for simplicity). The largest value you get is the maximum, the smallest value you get is the minimum.
A: An other thing you might try is to simplify the original function using trigonometric formulas. 
Recall the addition formula for the sine function: 
$$\sin(A+B)=\sin(A)\cdot\cos(B)+\sin(B)\cdot\cos(A).$$
Using this we see
$f(A)\cdot C=\sin(A)\cdot C+\cos(A)\cdot C=\sin(A+B)$, provided $\cos B =\sin B=C$. 
Next, we have $$1=\sin^2 B+\cos^2 B=2C^2$$
hence we get $$f(A)=\sqrt{2}\sin(A+B) \quad \text{ or }\quad  f(A)=-\sqrt{2}\sin(A+B)$$
which shows that the maximal possible value of $f$ is $\sqrt{2}$ and the minimal possible value is $-\sqrt{2}$. (Looking at the "circle-definition" of sine and cosine it is easy to see where these values are attained -- on the line $y=x$).
A: Let $$f(x)=\sin(x)+\cos(x)$$
Thus
$f(x)^2 = (\sin(x)+\cos(x))^2 = \sin(x)^2+2\sin(x)\cos(x)+\cos^2(x) = 1+\sin(2x)\Rightarrow$
$0\le f(x)^2 \le 2 \Rightarrow -\sqrt{2}\le f(x)\le\sqrt{2}$
How $f(\frac{\pi}{4})=\sqrt{2}$ and $f(\frac{5\pi}{4})=-\sqrt{2}$, then the maximum value is $\sqrt{2}$ and the minimum is $-\sqrt{2}$
