Finding Maxima/Minima of Trig Functions (Calculus)

Question

I'm having some trouble with this one, I have the following:

the cut off part is $\frac{9}{8}$

Now, I took the derivative of it.

$f'(x)=-4\sin(x)\cos(x)-cos(x)$

I factored it, equaled it to zero and got the following values:

$\cos(x)=0 => \frac {\pi}{2}$ or $\frac {3\pi}{2}$

$\sin(x)=\frac {-1}{4} =>$ some number (-0,25268....)

Now what I don't get is, how do I go from here to proving that $f(x)$ has a maximum of $\frac{9}{8}$.

Would appreciate some assistance, thanks in advance!

Answer 1

HINT

Recall that $$\cos(2x)=\cos^2 x - \sin^2 x =1-2\sin^2 x$$

and thus

$$f(x)=\cos (2x)-\sin x=1-2\sin^2 x-\sin x$$

then set $$t=\sin x\implies f(t)=-2t^2-t+1$$

and solve for $t\in[-1,1]$.