How to prove sin pi/6 = 1/2

I'm currently using Tom Apostol's 'Calculus'. One question involves proving sin pi/6 = 1/2 as well as the cos pi/6. The hint stated that I could use the previous exercise to help me which is showing the value of sin3x and cos3x. I tried using the fact that sin 2x= 2cosx sinx, so i could get sin6x = 2cos3x sin3x, but I don't know what to do after.


PS: I'm not looking for answers like, "draw a right triangle" and such. I need an analytical proof for the question.

So exactly how do you want this to be done? Using a multiple-angle formula to do this requires you to know the value of a sine/cosine of yet another value. For example, let's say you used the double-angle formula for sine:
sin(2x) = 2sin(x)cos(x).

With x = π/6, this becomes:
sin(π/3) = 2sin(π/6)cos(π/6).

At this point, the values of sin(π/3) and cos(π/6) need to be known.

The formula for sin(6x) will involve both sin(x) and cos(x) in it; there is no way to get sin(6x) in terms of sin(x) only unless we do:
sin(6x) = 2sin(3x)cos(3x) = 2sin(3x)√[1 - sin^2(3x)].

Using sin(3x) = 3sin(x) - 4sin^3(x), this becomes:
sin(6x) = 2[3sin(x) - 4sin^3(x)]√{1 - [3sin(x) - 4sin^3(x)]^2}.

Let x = π/6 here and you get:
sin(π) = 2[3sin(π/6) - 4sin^3(π/6)]√{1 - [3sin(π/6) - 4sin^3(π/6)]^2}
==> 2[3sin(π/6) - 4sin^3(π/6)]√{1 - [3sin(π/6) - 4sin^3(π/6)]^2} = 0.
(Note: in order to get this far, the assumption that sin(π) = 0 is required.)

Let y = sin(π/6) and this equation becomes:
2(3y - 4y^3)√[1 - (3y - 4y^3)^2] = 0.

By the zero-product property, this becomes:
3y - 4y^3 = 0 or √[1 - (3y - 4y^3)^2] = 0
==> y = -1, -√3/2, -1/2, 0, 1/2, and 1.

We know that y = sin(π/6) > 0 as π/6 is in Quadrant I, so this leaves us between:
y = 0, 1/2, and 1.

If you can use sin(0) = 0 and sin(π/2) = 1 (again, assuming that you know these values), then y = 1/2 is the only possible answer. Thus: