By simplifying the function to $-\sin(x)+\sin^2(x)$ I got a maximum and minimum of 2 and 0, using $x = \frac{3\pi}{2}$ and $x = \frac{\pi}{2}$ respectively. The answers are apparently 2 and $\frac{-1}{4}$, where exactly did I go wrong and are there other methods to solve this problem.
Maximum and minimum of $y = \cos(\frac{9}{2}\pi+x) + \sin^2(x)$
0
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trigonometry
2 Answers
5
HINT: $$\sin^2(x) - \sin(x) = \left(\sin(x) - \frac 12\right)^2 - \frac 14$$
4
Hint :
Did you try to compute the derivative ?
We have :
$$\frac{d}{dx}(-\sin(x)+\sin^2(x))=-\cos(x)+2\sin(x)\cos(x)=\cos(x)\left(2\sin(x)-1\right)$$
You should now be able to get the variations of your function and, in particular, to compute its max and min. ...