Hello, mad_munky838!

I can help you with the first one.

. . The solution is a strange one . . .

1) Express the following function as a single circular sinusoid:

. . f(x) .= .√3·sin(2x) + cos(2x)

Divide both sides by 2:

. . ˝·f(x) .= .˝·√3·sin(2x) + ˝·cos(2x)

We know that: .cos(30°) = ˝√3 .and .sin(30°) = ˝

So we have: .˝·f(x) .= .cos(30°)·sin(2x) + sin(30°)·cos(2x)

. . . . . - . . . - . . . . .= .sin(2x)·cos(30°) + cos(2x)·sin(30°)

From the compound-angle identity: .sin(A + B) .= .sin(A)·cos(B) + cos(A)·sin(B)

. . we have: .˝·f(x) .= .sin(2x + 30°)

Therefore: .f(x) .= .2·sin(2x + 30°)