# Thread: trigonometric identity

1. ## trigonometric identity

Prove that this equation is true.
sin22a-sin2a=sin3a x sina

2. I really don't understand your question. Do you mean

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

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

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

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

Please be more careful with your notation.

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

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

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

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

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

Please be more careful with your notation.
I want to mean : the first sin(22a)-sin(2a)=sin(3a)xsina

4. 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 $\displaystyle a$, so I thought you must have meant something else. Anyway, if we let $\displaystyle a=\frac{\pi}3$, we get

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

but

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

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

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

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

but

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

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

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

6. 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 $\displaystyle sin(na)$ 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