I dont even know where to start on this question!!

Evaluate: Sin(2Pi/3) and Cos(2Pi/3)

Printable View

- Oct 2nd 2007, 04:45 PMredpanda11Trig Helppppp Please!!!
I dont even know where to start on this question!!

Evaluate: Sin(2Pi/3) and Cos(2Pi/3) - Oct 2nd 2007, 04:49 PMJhevon
$\displaystyle \sin \frac {2 \pi}3 = \pm \sin \frac {\pi}3$ and $\displaystyle \cos \frac {2 \pi}3 = \pm \cos \frac {\pi}3$. those values you should know. but how do we figure out the signs? we do that by finding out what quadrant the angle $\displaystyle \frac {2 \pi}3$ is in.

Remember the mnemonic, All Students Take Calculus

if it is in the first quad: sine is positive and cosine is positive

if it is in the second quad: sine is positive and cosine is negative

if it is in the third quad: sine is negative and cosine is negative

if it is in the fourth quad: sine is negative and cosine is positive

so, what's the answer? - Oct 2nd 2007, 05:12 PMredpanda11
sin root 3/2,

cos -1/2? - Oct 2nd 2007, 05:23 PMredpanda11
sin + pi/3, cos -pi/3

- Oct 2nd 2007, 05:26 PMJhevon
what? why is there sine and cosine in your answer? the answers are just numbers.

please type coherently, i'm guessing you mean: $\displaystyle \sin \frac {2 \pi}3 = \frac {\sqrt {3}}2$ and $\displaystyle \cos \frac {2 \pi}3 = - \frac 12$

if so, that's correct

:confused: - Oct 2nd 2007, 05:30 PMredpanda11
sorry i meant root3/2

and -1/2 - Oct 2nd 2007, 05:32 PMredpanda11
sorry!! i understand it to a certain extent but what if it's like tan 19pi/6 how do you know what quadrant it is in??

- Oct 2nd 2007, 05:36 PMJhevon
you can keep adding or subtracting $\displaystyle 2 \pi$ if you don't like the size of the angle. you won't change it's value, you just change the number of revolutions to get there. so here, you would realize that $\displaystyle \frac {19 \pi}6 \equiv \frac {7 \pi}6$