1. ## Trigonometric equation

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

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

2. ## Re: Trigonometric equation

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

Then find value of $\displaystyle \sin (2006\; x)+\sin (2006\; y)$
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 $\displaystyle \cos x = \cos y$ and $\displaystyle \sin x = -\sin y$

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

4. ## Re: Trigonometric equation

Thanks Got it

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

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

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

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

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

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

5. ## Re: Trigonometric equation

Originally Posted by jacks
Thanks Got it

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

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

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

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

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

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

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

and so:

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

and the result follows immediately.

CB