It’s congruent. The line OL is shared which means it’s congruent for both triangles and since ON is congruent to ED, EN and OD have to be congruent.
When you are looking at a graph, a minimum point would be where the curve is decreasing, then begins to increase. Right at the point where it switches, the slope is a horizontal line, or 0. We can take the derivative is f(x), then look for all the x values where the slope (which is equal to the first derivative) is equal to zero.
f'(x) = 2 * -4sin(2x - pi)
The 2 comes from the derivative of the inside, 2x-pi.
So now set the derivative equal to 0.
-8sin(2x-pi) = 0
We can drop the -8 by dividing both sides by -8.
sin(2x-pi) = 0
This can be rewritten as arcsin(0) = 2x-pi
So when theta equals 0, what is the value of sin(theta)? At an angle of 0, there is just a horizontal line pointing to the right on the unit circle with length of 1. Sine is y/h, but there is no y value so it is just 0. If arcsin(0) = 0, we can now set 2x-pi = 0
2x = pi
x = pi/2
This is a critical number. To find the minimum value between 0 and pi, we need to find the y values for the endpoints and the critical number.
f(0) = -4
f(pi/2) = 4
f(pi) = -4
So the minimum points are at x=0 and x=pi
