sin(9pi/7) = -sin(2pi/7)

Stuck....

Step by step would be appreciated very much.

Thanks

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- Aug 30th 2012, 03:25 PMwaleedrabbaniexplain why these two expressions are equal
sin(9pi/7) = -sin(2pi/7)

Stuck....

Step by step would be appreciated very much.

Thanks - Aug 30th 2012, 05:19 PMskeeterRe: explain why these two expressions are equal
note ...

$\displaystyle \sin\left(\frac{9\pi}{7}\right) = \sin\left(\pi + \frac{2\pi}{7}\right)$

... now use the sum identity for sine - Aug 30th 2012, 05:37 PMrichard1234Re: explain why these two expressions are equal
If you draw the terminal angles $\displaystyle \frac{9\pi}{7}$ and $\displaystyle \frac{2\pi}{7}$ on a unit circle, you will see that they differ by $\displaystyle \pi$. The sine of the angle is the y-coordinate of that angle/point on the unit circle. You will see that one is the negative of the other.

In general, $\displaystyle \sin (x + \pi) = -\sin x$. - Aug 30th 2012, 07:38 PMProve ItRe: explain why these two expressions are equal
- Aug 31st 2012, 03:11 AMskeeterRe: explain why these two expressions are equal
- Aug 31st 2012, 03:12 AMProve ItRe: explain why these two expressions are equal
- Aug 31st 2012, 10:36 AMskeeterRe: explain why these two expressions are equal
- Sep 7th 2012, 08:57 PMtrutinhRe: explain why these two expressions are equal
i am appreciate you ... this is right .... work smart not hard...

\sin\left(\frac{9\pi}{7}\right) = \sin\left(\pi + \frac{2\pi}{7}\right)