Finding Exact Values

• Dec 3rd 2008, 03:26 PM
shiel8
Finding Exact Values
I have 3 problems that I need help with. All of them are 'give the exact value of the trig function'. I am totally lost here.

cos 11pi/4

sin 7pi/6

tan 5pi/3
• Dec 3rd 2008, 04:37 PM
skeeter
these are all angles on the unit circle.

• Dec 4th 2008, 06:26 PM
nzmathman
$\displaystyle \cos{\left(\frac{11\pi}{4}\right)}$

$\displaystyle = \cos{\left(3\pi - \frac{\pi}{4}\right)}$

$\displaystyle = \cos{(3\pi)}\,\cos{\left(\frac{\pi}{4}\right)} + \sin{(3\pi)}\,\sin{\left(\frac{\pi}{4}\right)}$

We know $\displaystyle \sin{(n\pi)} = 0 , n\in \mathbb{I}$
And that $\displaystyle \cos{(3\pi)} = -1$

So, $\displaystyle \cos{(3\pi)}\,\cos{\left(\frac{\pi}{4}\right)} + \sin{(3\pi)}\,\sin{\left(\frac{\pi}{4}\right)}$

$\displaystyle = -\cos{\left(\frac{\pi}{4}\right)} = -\frac{\sqrt{2}}{2}$

$\displaystyle \sin{\left(\frac{7\pi}{6} \right)} = \sin{\left(\pi + \frac{\pi}{6}\right)}$

$\displaystyle = \sin{(\pi)}\,\cos{\left(\frac{\pi}{6}\right)} + \cos{(\pi)}\,\sin{\left(\frac{\pi}{6}\right)}$

$\displaystyle = -sin{\left(\frac{\pi}{6}\right)} = -\frac{1}{2}$

$\displaystyle \tan{\left(\frac{5\pi}{3}\right)} = \tan{\left(2\pi - \frac{\pi}{3}\right)}$

$\displaystyle = \frac{\tan{(2\pi)} - \tan{\left(\frac{\pi}{3}\right)}}{1 + \tan{(2\pi)}\,\tan{\left(\frac{\pi}{3}\right)}}$

$\displaystyle =-\tan{\left(\frac{\pi}{3}\right)} = -\sqrt{3}$
• Dec 5th 2008, 06:25 AM
shiel8
Thanks nzmathman. The walkthrough of the steps is very helpful.