Finding Maximum/Minimum of Trigonometric Functions I have a question with regards to finding the maximum and minimum, points of Trigonometric Functions. I so far understand that I have to take the first derivative and set the expression equal to $0$ to determine the values, but with trigonometric functions, I'm not sure how to determine it's value or rather if I am doing it correctly.
Would the following be correct?
$$y=sin(x) + cos (x), 0\le \pi \le 2\pi$$
$$y' =cos(x) - sin(x)$$
$$0 = cos(x) - sin(x)$$
From here on out do I set each individual, $cos(x)$ and sin(x), equal to 0?
Then solve for the values?
If so, values that would satisfy $cos(x)= 0$ would be $x= \frac{\pi}{2}, \frac{3\pi}{2}$ and $sin(x) = 0$ would be $x=\pi, 2\pi$
Thank you for you all in advance for helping me.
 A: One  short way might be $$f(x)=\sin { \left( x \right)  } +\cos { \left( x \right)  } =\sqrt { 2 } \left( \frac { \sqrt { 2 }  }{ 2 } \sin { \left( x \right) +\frac { \sqrt { 2 }  }{ 2 } \cos { \left( x \right)  }  }  \right) =\sqrt { 2 } \sin { \left( x+\frac { \pi  }{ 4 }  \right)  }  $$
as $1\le \sin { \left( x+\frac { \pi  }{ 4 }  \right) \le 1 } $

$$-\sqrt { 2 } \le \sqrt { 2 } \sin { \left( x+\frac { \pi  }{ 4 }  \right)  } \le \sqrt { 2 } \\ $$

A: No, you need to find $x$ that makes $\cos x - \sin x = 0$.  There are trig identities (which is why trig is a prereq for calculus) that you can use to change the expression into a single trig function.  Or you can write $\cos x = \sin x$ and hence $\tan x = 1$.  So $x= \pi/4,$ or $5\pi/4$ plus any multiple of $2\pi$ you like.
A: You must find those $x$ such that $y'(x)=0$, i.e., $\cos x - \sin x =0 \implies \cos x = \sin x$. In the given domain, these guys are $\pi/4, 5\pi/4$. You can see this by the following way: $\cos x=\sin x \implies 2 \cos^2 x = 1 \implies |\cos x|=\sqrt{2}/2$. So, if $x$ satisfies the condition, then it should be one of the above. Furthermore, the values given above satisfies the condition $\cos x=\sin x$.
