Let's first convert #sin2x# into something a bit more friendly to work with. There is an identity that states:
#sin2x=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 #cosx# so we only need to find #sinx#. We can do that by realizing that #cosx="adj"/"hyp"=1/2# and from the Pythagorean Theorem:
#a^2+b^2=c^2#
#1^2+b^2=2^2#
#1+b^2=4#
#b^2=3#
#b=sqrt3#
(We could also have known this by remembering about the 30, 60, 90 triangle).
We can now see that #sinx=sqrt3/2#.
Now let's work through the quadrants.
When #cosx# is positive, it can be in the first or fourth quadrant. Now, in the first quadrant, #sinx# is also positive, but in the fourth quadrant, #sinx# is negative. This means that when #cosx=1/2#, #sinx=+-sqrt3/2#.
We can then substitute in:
#sin2x=2(+-sqrt3/2)(1/2)=+-sqrt3/2#
