Analytic Trig Problems

I just have no idea how to do these following problems.

Use the half angle formulas to simplify the expression square root[(1-cos6x)/(2)]

Find the exact zeros of the function in the interval [0, 2pi)

f(x) = sin (x/2) - cosx
h(x) = cos (x/2) - sinx

Please help!

1. Use the half-angle sine formula:

sin(t/2) = √[(1 - cos t) / 2]

If you let t=6x then the right side is the same as your expression above. The left side, which is equivalent, becomes:

sin(t/2) = sin(6x/2) = sin(3x)

...so sin(3x) is the simplification. This is only good when sin(t/2) >= 0, because that's the limit of the half-angle formula in square-root form. The original squared version:

sin²(t/2) = [1 - cos t]/2

gives the full story:

|sin(t/2)| is equivalent to instead, with t=will get you to the more complete:

√[(1 - cos(6x)) / 2] = |sin 3x|

...which is good for all values of x.

Double-angle cosine formulas work better for the second problem. Let x=2t and solve:
sin t - cos (2t) = 0
cos 2t = sin t
1 - 2 sin² t = sin t
2 sin² t + sin t - 1 = 0

(2 sin t - 1)(sin t + 1) = 0
sin t = 1/2 ....or sin t = -1

Since 0 <= x = 2t < 2π, then 0 <= t < π, so solve those two inverse sine problems for t in [0,π). Since sin t >= 0 in this interval, only the arcsin(1/2) solutions occur. Both are solutions. x is in{ π/3, 5π/3 }.

The other problem looks easier, since there's no quadratic to factor. cos t = sin (2t) = 2 sin t cos t is solved when:

(2 sin t - 1)(cos t) = 0

...or when cos t = 0 (at t=π/2, x=π) or when sin t = 1/2, which is the same two points found above.