Math Help - Trigonometric equation

1. Trigonometric equation

If $\cos x = \cos y$ and $\sin x = -\sin y$

Then find value of $\sin (2006\; x)+\sin (2006\; y)$

2. Re: Trigonometric equation

Originally Posted by jacks
If $\cos x = \cos y$ and $\sin x = -\sin y$

Then find value of $\sin (2006\; x)+\sin (2006\; y)$
So what have you tried?

Clearly y=-x satisfies the condition, but does anything else?

CB

3. Re: Trigonometric equation

Originally Posted by jacks
If $\cos x = \cos y$ and $\sin x = -\sin y$

Then find value of $\sin (2006\; x)+\sin (2006\; y)$
Are x and y mutually exclusive?

4. Re: Trigonometric equation

Thanks Got it

Let $z=\cos x+i\sin x$ and $\omega = \cos y +i\sin y$

Then from Given equation $z=\bar{\omega}$

So $\cos (2006 \;x)+i\sin (2006\; x) = z^{2006}$

OR $Im(z^{2006}) = \sin (2006\; x)$

and $Im(\omega^{2006}) = \sin (2006\; y)$

So $\sin (2006\; x)+\sin (2006\; y) = Im\left(z^{2006}+\omega^{2006}\right)=Im\left(\bar {\omega})^{2006}+\omega^{2006}\right) = 0$

5. Re: Trigonometric equation

Originally Posted by jacks
Thanks Got it

Let $z=\cos x+i\sin x$ and $\omega = \cos y +i\sin y$

Then from Given equation $z=\bar{\omega}$

So $\cos (2006 \;x)+i\sin (2006\; x) = z^{2006}$

OR $Im(z^{2006}) = \sin (2006\; x)$

and $Im(\omega^{2006}) = \sin (2006\; y)$

So $\sin (2006\; x)+\sin (2006\; y) = Im\left(z^{2006}+\omega^{2006}\right)=Im\left(\bar {\omega})^{2006}+\omega^{2006}\right) = 0$
If you had indicated that you were familiar with complex analysis we could have pointed out that the conditions give:

$e^{ix}=e^{-iy + 2n\pi i}, \ \ n=...-1, 0, 1, ...$

and so:

$e^{2006 ix}=e^{- 2006 iy} e^{4012 \pi i}=e^{- 2006 iy}$

and the result follows immediately.

CB