triple angle identity

Question

sin(3x) = 3sin(x) - 4sin³(x)

which is also equal to

sin³(x) = [3sin(x) - sin(3x)] / 4

For this equation, if the problem states sin^3(theta+120), why would I only substitute the value inside the parentheses into theta next to 3sin but not in -sin(3theta).

For example why is below true

sin³(x+120) = [3sin(x+120) - sin(3x)] / 4

but not below?

sin³(x+120) = [3sin(x+120) - sin(3(x+120))] / 4

Answer 1

Actually your answer is also correct!

Your answer is:

sin³(x+120) = [3sin(x+120) - sin(3(x+120))] / 4

But notice that this is the same as

sin³(x+120) = [3sin(x+120) - sin(3x+360)] / 4

But then 360 is the period of sine. So sin(3x+360) is just the same thing as sin(3x). This turns your answer into the other answer that you were asking about.

Answer 2

How do we get there? Let's have a look.

sin 3x = sin (2x + x)

sin 3x = sin 2x cos x + cos 2x sin x

sin 3x = 2 sin x cos x cos x + (1 − 2 sin²x) sin x

sin 3x = 2 sin x (1 − sin²x) + sin x − 2 sin³x

sin 3x = 3 sin x − 4 sin³x

Transforming,

sin³x = 3 sin x − sin 3x / 4

here angles are different, one sin represents the angle 3x while another represents x.

We know that for a function sin bx, the period is 2π/|b| which implies the period of sin 3x is 2π/3.

Hence, the graph of sin 3x is narrower than the graph of sin x as the period of sin3x is one-third the period of sin x (Period of sin x is 2π)

Now, let us plot the graph of sin 3x by taking some points on the graph and joining them.

Let us consider a few points for y = sin 3x and y = sin x and plot them.

1. When x = 0, 3x = 0 ⇒ sin x = 0, sin3x = 0
2. When x = -π/6, 3x = -π/2 ⇒ sin x = -1/2, sin3x = -1
3. When x = π/6, 3x = π/2 ⇒ sin x = 1/2, sin3x = 1
4. When x = π/2, 3x = 3π/2 ⇒ sin x = 1, sin3x = -1
5. When x = -π/2, 3x = -3π/2 ⇒ sin x = -1, sin3x = 1