$$2\sin(45^\circ-x)\cos(45^\circ-x)$$
I know you have to use the double angle formula for sine, but what next?
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I know you have to use the double angle formula for sine, but what next?
Using the double angle formula for sine, $$2\sin(\theta)\cos(\theta)=\sin(2\theta),$$ we see that $$2\sin(45-x)\cos(45-x)=\sin(2\cdot(45-x))=\sin(90-2x)$$ Now use the identity that allows you to simplify $\sin(90-\theta)$, where in our case we have $\theta=2x$. You'll then want to use the double angle formula for cosine.
Hint : use $$2 \sin a \cdot \cos a = \sin 2a$$ and $$\sin(90-a)=\cos a$$
double angle formula for sine,
$2sin(\theta)cos(\theta)=sin(2\theta)$,
now,
$2sin(45−x)cos(45−x)=sin(2⋅(45−x))=sin(90−2x)$.
$sin(90-2x)=cos2x$
since,$(90-\theta)=cos(\theta)$, since cosine is positive in first and fourth quadrant, and maximum value of $\theta$ can not exceed $180^o$.
Now, Its easy ahead