Let's first convert sin2xsin2x into something a bit more friendly to work with. There is an identity that states:
sin2x=2sinxcosxsin2x=2sinxcosx
For this next bit, I'm going to focus on Quadrant 1 so that we can figure out the lengths of the sides. Once we have that, then we'll worry about other quadrants.
We're given cosxcosx so we only need to find sinxsinx. We can do that by realizing that cosx="adj"/"hyp"=1/2cosx=adjhyp=12 and from the Pythagorean Theorem:
a^2+b^2=c^2a2+b2=c2
1^2+b^2=2^212+b2=22
1+b^2=41+b2=4
b^2=3b2=3
b=sqrt3b=√3
(We could also have known this by remembering about the 30, 60, 90 triangle).
We can now see that sinx=sqrt3/2sinx=√32.
Now let's work through the quadrants.
When cosxcosx is positive, it can be in the first or fourth quadrant. Now, in the first quadrant, sinxsinx is also positive, but in the fourth quadrant, sinxsinx is negative. This means that when cosx=1/2cosx=12, sinx=+-sqrt3/2sinx=±√32.
We can then substitute in:
sin2x=2(+-sqrt3/2)(1/2)=+-sqrt3/2sin2x=2(±√32)(12)=±√32
