I understand basic $\sin x$ and $\cos x$ min/max, but I am having a problem solving the minimum and maximum of the following:
$f(x) = \sin^2 x – \sin x$
Oh, and the range is $0 \le x \le \frac{3\pi}{2}$
[Math] How to find the maximum and minimum of a sinusoidal function
trigonometry
Best Answer
$$\begin{align*} f(x) &= \sin^2 x - \sin x \\ &= \sin^2 x - 2 \cdot \tfrac{1}{2} \sin x + (\tfrac{1}{2})^2 - \tfrac{1}{4} \\ &= \left( \sin x - \tfrac{1}{2} \right)^2 - \tfrac{1}{4}\end{align*}.$$ Because the square of a real number is nonnegative, $f$ attains a minimum if $\sin x = \frac{1}{2}$, and the consequences are straightforward.
To determine the maximum value, observe that $|\sin x| \le 1$; consequently, $f$ is maximized if $(\sin x - \tfrac{1}{2})^2$ is made as large as possible. By inspection, this occurs if $\sin x = -1$.