# Can I simplify $\cos \frac{1234 \pi}{5}$ and $\sin \frac{1234 \pi}{5}$

I have

$$\cos \frac{1234 \pi}{5} + i \cdot \sin \frac{1234 \pi}{5}$$

I believe I can simplify the $1234$ further, but how?

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$1234=246\cdot 5 + 4$ –  Blah Feb 26 '12 at 8:11

$\frac{1234\pi}5 = 246\pi + \frac{4\pi}5$

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To complement mrf, we can say that $cos(\frac{1234 \pi}{5})=cos(\frac{4 \pi}{5}$). Because there are the periodic identities which stays that:

$\sin( \theta+ 2 \pi n)= \sin \theta$

$\cos( \theta+ 2 \pi n)= \cos \theta \qquad n \in \mathbb{Z}$

This happens because the period of sine and cosine funtions is $2 \pi$.

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You can not say that $\frac{1234\pi}{5} = \frac{4\pi}{5}$; you can just say that their sines (and cosines) are equal. –  Steven Stadnicki Feb 27 '12 at 23:43
Rather than saying these fractions are equal, one should say that they are congruent mod $2\pi$. –  Michael Hardy Feb 27 '12 at 23:49
Steven Stadnicki you are right.I forgot to fit in the question.Its done –  João Feb 28 '12 at 0:16