# Could anyone explain how to simplify $2\sin(45^\circ-x)\cos(45^\circ-x)$?

$$2\sin(45^\circ-x)\cos(45^\circ-x)$$

I know you have to use the double angle formula for sine, but what next?

• Yeah, I got up to that, but the answer states cos 2x. How do you expand sin 2(45-x)? Apr 8, 2013 at 8:33

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.

• Thanks! That cleared things up! Apr 8, 2013 at 8:37
• Glad I could help! Apr 8, 2013 at 8:37

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 $$180^o$$.