OK, in this problem we want to find F of H of Pi 6, so we're going to actually put Pi 6 into the H equation. Remember, when we have lots of parentheses, we always do the innermost first. So I'm going to stick PI6 in every time I see the unknown in the H equation. So if I have H of X equal to X, now I'm going to have H of Pi 6 equal 2 * π six. Remember 2 is really 2 / 1. So the two and the six can reduce and we get π thirds. So this is really saying H of Pi 6 is equivalent to π thirds. So now when I have F of H of Pi 6:00, I'm going to take out this H of Pi 6 and put in what it equaled, which was π thirds. So now in the F equation, the F of X equals sine X. Every time I see that X unknown, I'm going to put in π thirds. So now I want sine of π thirds. So if we think about our unit circle π thirds is equivalent to 60°. So it's a 3060 right triangle. So if we thought about the hypotenuse being two, so that if we had an equilateral triangle here, that would split the space up so that each of those halves would be one. Using the identity, Pythagorean identity, we could say a ^2 + b ^2 = C ^2, 1 ^2 + b ^2 equal 2 ^2. Solving that out, we can see B is root 3, and then because it's a unit circle, the hypotenuse has to be one. So I'm going to divide all three sides by two. So our cosine, how far over we went was 1/2. How far up we went was root 3 / 2 and our tangent is sine divided by cosine or y / X so root 3. So we wanted sine of π thirds. So our final answer sine of π thirds is how far up we went in the unit circle at that angle, so root 3 / 2.