Question

If `cosx=1/2`, how do you find `sin2x`?

Answer 1

`sin2x=2sinxcosx=2(sqrt3/2)(1/2)=+-sqrt3/2`

Explanation:

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`