# Find all critical numbers of $f(\theta) =\cos(2\theta), 0 < \theta < \frac{\pi}{4}$ then use first derivative test to find rel max or min

The theta and cosine are really throwing me off. I had one of my friends help me out, but the work really does not make sense to me. They got $\sin(0) = 0$ is the only critical number and there is no min/max.

Thank you very much!

• Are you using en.wikipedia.org/wiki/Second_derivative_test – lab bhattacharjee Jun 22 '15 at 3:26
• Well, can you show their work? We may be able to help you make sense of it, or confirm that it makes no sense. You could also post your own attempts and thoughts. Those will help us gauge your experience level, and better tailor answers to your needs! – Cameron Buie Jun 22 '15 at 3:28

One could approach it through the Chebyshev Polynomials of the first kind. Since $T_2(x) = 2x^2 - 1$ or $2\cos^2(x) - 1 = \cos(2x)$, we only need to find the $t$-values of the critical points of $2t^2 - 1$ in the domain $[-1,1]$ and find their corresponding $x$-values if we let $t = cos(x)$ (the reason we restrict the domain is because $cos(x)$ can only take values in $[-1,1]$). The only critical points are $t=0$ which is a minimum, and $t=\pm 1$ which are maximums. Solving for $x$ in the domains for $x$ given shows us that the only maximums and minimums occur at the endpoints of the domain given.
suppose you derive $$f(x)=\cos(2\theta)\Rightarrow f^\prime(\theta)=-2\sin(2\theta)$$now: $$f^\prime(x)=0\iff\sin2\theta=0\iff\theta=\pi k \quad(k\in\mathbb{Z})$$but your range is $[0,\frac \pi 4]$ so the only extrmum is indeed at $x=0$. by simple substitution in the derivative (or simply since the $\cos$ is monotonically decreasing on this interval), you get it's a maximum point.
EDIT: One may also consider $x=\frac \pi 4$ as a minimum point on this interval (since exists left neighborhood with greater values and it's the right edge of the interval).
• See my edit, $k\in\mathbb{Z}$ – Alan Watts Jun 22 '15 at 3:41