neeeed help plz//trigonometry?

Question

Note that the point value of each question is given in parentheses

6. (5) Derive the identity for sin 3x in terms of sin x

7. (5) Using the double angle formula, find sin 120 degrees

Answer 1

6. To solve for sin(3x) in terms of sin(x), all you have to do is keep in mind that

sin(3x) = sin(2x + x)

Now, apply the sine addition identity.

sin(2x + x) = sin(2x)cos(x) + sin(x)cos(2x)

Now, apply the double angle identity. A reminder:
sin(2x) = 2sin(x)cos(x)
cos(2x) = cos^2(x) - sin^2(x)

2sin(x)cos(x)cos(x) + sin(x) [cos^2(x) - sin^2(x)]

Now, expand

2sin(x)cos^2(x) + sin(x)cos^2(x) - sin^3(x)

We could, in theory, change everything to sines.

2sin(x)[1 - sin^2(x)] + sin(x)[1 - sin^2(x)] - sin^3(x)

2sin(x) - 2sin^3(x) + sin(x) - sin^3(x) - sin^3(x)

3sin(x) - 4sin^3(x)

7.

sin(120) = sin(2 * 60)
= 2sin(60)cos(60)
= 2[sqrt(3)/2] [1/2]
sqrt(3)/2

Answer 2

6.
cos3x + isin3x = (cosx +i sinx)^3
sin3x = 3cos^2 x sinx-sin ^3 x
= 3sinx-4sin^3 x

7.
sin 120
= sin2*60
= 2sin60cos60
= sin60
= √3 / 2