How do u find critical points for trigonometric derivatives?

Question

Say the derivative is sin x cos x. When u set it equal to zero, how do u solve it? sin x cos x = 0 How would u find the min max values and concavity (2nd derivative) and inflection points?

Answer 1

If f(x) = (1/2) sin^2(x), then
f'(x) = (1/2) 2sin(x) cos(x), or
f'(x) = sin(x)cos(x)

When setting this value to 0, we get

0 = sin(x)cos(x)

Therefore, to solve this, we equate each factor to 0.
sin(x) = 0, cos(x) =0

To obtain the solution to sin(x) = 0, we have to ask ourselves: Where on the graph is sine equal to 0? The answer to that question is at 0 and pi. Note that there are no interval restrictions, so the actual answer is
x = { k*pi | k is an integer }, or 0, pi, 2pi, 3pi, 4pi, etc...

cos(x) = 0 when x = pi/2, 3pi/2 on a restricted interval. Therefore, the general solution is
x = {pi/2 + kpi | k is an integer}, or pi/2, 3pi/2, 5pi/2, 7pi/2, etc.

We can merge those two solutions into one.

x = {k*pi/2 | k is an integer}, which will cover 0, pi/2, pi, 3pi/2, 2pi, and so forth.

So we have an infinite number of critical points. We would test
f(0), f(pi/2), f(pi) to determine the local min and max, and realize it's periodic.

The second derivative is calculated using the following:

f'(x) = sin(x)cos(x)
f''(x) = cos(x)cos(x) + sin(x)(-sin(x)
f''(x) = cos^2(x) - sin^2(x) = cos(2x)

We find the points of concavity by solving cos(2x) = 0
2x = {pi/2 + 2k*pi | k an integer }
x = {pi/4 + k*pi | k an integer }

Answer 2

You have to find the period of it and then once you find a single solution, it is every multiple of the period to the right and left of it.

As far as solving it, maybe graph it? Or use a TI-89... :)

Answer 3

You and your math teacher need to have a LONG talk.