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

[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.

 

[haruspex]

you could divide the 4 into the -8

Seems reasonable. You could also consider 2x as y, simplifying the innards.

sin(2x+2x) can be written as: 2Sin(x)Cos(x) + 2Sin(x)Cos(x)

No.

 

[Tyrion101]

What do I do with the 2?

 

[haruspex]

What do I do with the 2?

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

 

[Tyrion101]

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

 

[haruspex]

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

How would you expand sin(2x)?

 

[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.

 

[Mark44]

am I correct in assuming that taking just sin(2x+2x) can be written as: 2Sin(x)Cos(x) + 2Sin(x)Cos(x)?

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)?

 

[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.

 

[Maged Saeed]

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.

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?

 

[SteamKing]

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.

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).

 

[haruspex]

I believe it is 2Sin(x)Cos(x) but you said that that was not how you did 4x

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)).