Reducing fractions with pi in them

**Ragnarok** · January 15th 2013, 03:38 PM

I am trying to simplify the expression

$\cos\frac{35\pi}{4}+i\sin\frac{35\pi}{4}$

The cosine and sine functions can be reduced modulo $2\pi$ , right? So I wrote this as

$\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}$

after subtracting $\frac{32\pi}{4}=8\pi$ from each of the arguments. But checking this equality with my calculator I get something different. What am I doing wrong?

**russo** · January 15th 2013, 04:02 PM

Maybe something's wrong with your calculator settings. I get $cos\left(\frac{35\pi}{4}\right) = cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ and $sin\left(\frac{35\pi}{4}\right) = sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$

**topsquark** · January 15th 2013, 04:06 PM

A common mistake might be at play. Make sure you are in 'radian' mode.

-Dan

**Ragnarok** · January 15th 2013, 04:32 PM

Ah, I should have known! Thanks!

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