Trigonometric equation

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December 31st 2011, 11:46 PM
jacks

Trigonometric equation

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

Then find value of $\sin (2006\; x)+\sin (2006\; y)$
January 1st 2012, 12:32 AM
CaptainBlack

Re: Trigonometric equation

Quote:

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
January 1st 2012, 01:54 AM
Prove It

Re: Trigonometric equation

Quote:

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?
January 1st 2012, 05:26 AM
jacks

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 $\sin (2006\; x)+\sin (2006\; y) = Im\left(z^{2006}+\omega^{2006}\right)=Im\left(\bar {\omega})^{2006}+\omega^{2006}\right) = 0$
January 1st 2012, 06:50 AM
CaptainBlack

Re: Trigonometric equation

Quote:

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