# 4Sine(4X) = -8Sin(2x) Double angle identity

1. Mar 4, 2015

### Tyrion101

I kind of understand what to do with this when there are no numbers in front of the expressions, I also kind of understand that you can rewrite 4Sine(4x) as 4Sine(2x+2x) hat do I do with the 4 and 8? In an algebra problem you could divide the 4 into the -8, then simplify that expression, am I correct in assuming that taking just sin(2x+2x) can be written as: 2Sin(x)Cos(x) + 2Sin(x)Cos(x)? Or am I way off base in all of my assumptions? The explanation in my math book doesn't explain the whole problem.

2. Mar 4, 2015

### haruspex

Seems reasonable. You could also consider 2x as y, simplifying the innards.
No.

3. Mar 4, 2015

### Tyrion101

What do I do with the 2?

4. Mar 4, 2015

### haruspex

You mean after dividing the 4 in to the 8? Leave it be for now. Concentrate on expanding the sin(4x).

5. Mar 4, 2015

### Tyrion101

Expanding the (4x) this is the bit je ne comprends pas. (I don't understand)

6. Mar 4, 2015

### haruspex

How would you expand sin(2x)?

7. Mar 4, 2015

### Tyrion101

I believe it is 2Sin(x)Cos(x) but you said that that was not how you did 4x, so now I'm thinking: Sin(2x)Cos(2x) Or it could be 2SinCos but that doesn't seem right. If the middle one is correct I think I may understand now.

8. Mar 4, 2015

### Staff: Mentor

No, and here's why. You cannot split up sin(2x + 2x) to sin(2x) + sin(2x) as if "sin" were multiplying (2x + 2x). You are confusing function notation with the distributive law, which says that a(b + c) = ab + ac.

sin(2x + 2x) is no more equal to sin(2x) + sin(2x) than is $\sqrt{a + b}$ the same as $\sqrt{a} + \sqrt{b}$.

The gist of this problem, at least according to what you wrote in the thread title, is solve the equation 4sin(4x) = -8sin(2x).

Can you solve the equation 4sin(2y) = -8sin(y)?

Last edited: Mar 5, 2015
9. Mar 5, 2015

### Tyrion101

Then I have absolutely no clue what is going on with that part of the problem, I know it relates somehow to the sum and difference identities, but the two with the x confuses me.

10. Mar 5, 2015

### Maged Saeed

I think Mr @Mark44 is true.
Can you start solving this equation?
The only identity you need is
sin(2u)=2sin(u)cos(u)
What is the first step now?

11. Mar 5, 2015

### SteamKing

Staff Emeritus
The trigonometric identities for finding the sine and cosine of the sum of two angles are:

sin (α + β) = sin (α) * cos (β) + cos (α) * sin (β)
cos (α + β) = cos (α) * cos (β) - sin (α) * sin (β)

If you want to expand sin (2x) = sin (x + x), which implies that α = β = x, then

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

sin (4x) = sin (2x + 2x), which implies α = β = 2x

I'll leave it to you to work out the rest of the expansion for sin (4x).

12. Mar 5, 2015

### haruspex

Yes, it's 2sin(x)cos(x), and you can do 4x 'that way', but the way you extended it to 4x was wrong. Think of it as sin(2(2x)).

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