CI: Find max/min values for f(x)=2sin(x)cos(x) on the interval [0,pi]

This is a Calculus I problem I need help with, am I terrible with trigonometric identities. Could someone please help me step by step, starting with finding the derivative? It would be greatly appreciated. Thanks for any help:

Find the maximal and minimal values for the function f(x)=2sin(x)cos(x) on the interval [0,pi]

I'd simplify this first:

f(x) = 2 * sin(x) * cos(x)
f(x) = sin(2x)

That's the double angle identity

Now, take the derivative and find when f'(x) = 0

f(x) = sin(2x)
f'(x) = 2 * cos(2x)
f'(x) = 0
0 = 2 * cos(2x)
0 = cos(2x)
pi/2 , 3pi/2 , 5pi/2 , 7pi/2 = 2x
pi/4 , 3pi/4 , 5pi/4 , 7pi/4 = x

But when do the maximums occur and when do the minimums occur?

f(x) = sin(2x)

sin(2 * pi/4) , sin(2 * 3pi/4) , sin(2 * 5pi/4) , sin(2 * 7pi/4) =>
sin(pi/2) , sin(3pi/2) , sin(5pi/2) , sin(7pi/2) =>
1 , -1 , 1 , -1

Maximums occur when x = pi/4 and 5pi/4
Minimums occur when x = 3pi/4 and 7pi/4

Also, I just noticed that the domain was from 0 to pi (I don't know why, but I thought it was 0 to 2pi)

Max: x = pi/4
Min: x = 3pi/4

1. Differentiate and set derivative to zero:
dy/dx=2cos^2(x)-2sin^2(x)=0

2. Take the factor of 2 over to the rhs:
cos^2(x)-sin^2(x)=0

3. substitute sin^2(x)=1-cos^2(x):
cos^2(x)-(1-cos^2(x))=0
2cos^2(x)-1=0

4. Solve for x:
cos(x)= (+/-)sqrt(1/2)

5. And your two solution in the interval [0,pi] are:
x=acos(sqrt(1/2)), pi-acos(sqrt(1/2))
=pi/4, 3pi/4

2sin(x)cos(x) = sin(2x)

that makes the derivative:

(sin(2x))' = 2cos(2x)

2cos(2x) = 0 when 2x = pi/2 or 2x = 3pi / 2

which means:

x = pi/4 or x = 3pi/4

f(pi/4) = sin(2pi/4) = sin( pi/2) = 1

f(3pi/4) = sin(6pi/4) = sin(3pi/2) = -1

maximal value is 1 at x=pi/4
minimal value is -1 at x=3pi/4