# trigonometric identity

• May 4th 2008, 09:30 AM
blertta
trigonometric identity
Prove that this equation is true.
sin22a-sin2a=sin3a x sina
• May 4th 2008, 09:48 AM
Reckoner
I really don't understand your question. Do you mean

$\sin(22a)-\sin(2a)=\sin(3a)\sin a$?

$\sin^2(2a)-\sin^2a=\sin^3a\sin a$?

$\sin^{22}a-\sin^2a=\sin^3a\sin a$?

$\sin^2(2a)-\sin(2a)=\sin(3a)\sin a$?

• May 4th 2008, 01:27 PM
blertta
Quote:

Originally Posted by Reckoner
I really don't understand your question. Do you mean

$\sin(22a)-\sin(2a)=\sin(3a)\sin a$?

$\sin^2(2a)-\sin^2a=\sin^3a\sin a$?

$\sin^{22}a-\sin^2a=\sin^3a\sin a$?

$\sin^2(2a)-\sin(2a)=\sin(3a)\sin a$?

I want to mean : the first sin(22a)-sin(2a)=sin(3a)xsina
• May 4th 2008, 01:42 PM
Reckoner
Quote:

Originally Posted by blertta
I want to mean : the first sin(22a)-sin(2a)=sin(3a)xsina

Sorry. This equation isn't true for all $a$, so I thought you must have meant something else. Anyway, if we let $a=\frac{\pi}3$, we get

$\sin(22a)-\sin(2a)=-\frac{\sqrt{3}}2$

but

$\sin(3a)\sin a = 0$.

Was there supposed to be a restriction on $a$?
• May 5th 2008, 03:23 AM
blertta
Quote:

Originally Posted by Reckoner
Sorry. This equation isn't true for all $a$, so I thought you must have meant something else. Anyway, if we let $a=\frac{\pi}3$, we get

$\sin(22a)-\sin(2a)=-\frac{\sqrt{3}}2$

but

$\sin(3a)\sin a = 0$.

Was there supposed to be a restriction on $a$?

The question is: Prove that the equation is true for every allowable value of the variable.
Hope you understand me.
Sorry for my english.
• May 5th 2008, 03:42 AM
topsquark
Quote:

Originally Posted by blertta
The question is: Prove that the equation is true for every allowable value of the variable.
Hope you understand me.
Sorry for my english.

Then you want to solve the equation for a.

My calculator comes up with 18 solutions. I would not be surprised if there were actually 22 possibles (thus leaving 4 complex solutions.) Generally an equation with $sin(na)$ in it will reduce to an nth degree polynomial so you would be stuck solving a 22nd degree polynomial equation, which is impossible to do in general. You are going to be stuck doing numerical approximations.

However notice that there is at least one nice solution: a = 0.

-Dan