# R.O.C of trig functions

• Jan 18th 2011, 09:11 PM
TN17
R.O.C of trig functions
Determine the average rate of change of the function y = 2cos (x - pi/3) + 1 for the following interval:
pi/2 < x < 5pi/4

AROC = [ f(x2)- f(x1) ] / x2- x1

For an approx. value, I would set the calculator in radian mode and just plug everything into the equation, right?
Where I'm stuck is how to get an EXACT answer.

I'm not sure if using the new trig identities will help.
• Jan 18th 2011, 10:46 PM
Unknown008
$2\cos\left(\dfrac{5\pi}{4}- \dfrac{\pi}{3}\right) + 1 = 2\left(\cos\left(\dfrac{5\pi}{4}\right)\cos\left(\ dfrac{\pi}{3}\right) + \sin\left(\dfrac{5\pi}{4}\right)\sin\left(\dfrac{\ pi}{3}\right) \right) + 1$

$2\cos\left(\dfrac{\pi}{2}- \dfrac{\pi}{3}\right) + 1 = 2\left(\cos\left(\dfrac{\pi}{2}\right)\cos\left(\d frac{\pi}{3}\right) + \sin\left(\dfrac{\pi}{2}\right)\sin\left(\dfrac{\p i}{3}\right) \right)+1$

$\cos\left(\dfrac{5\pi}{4}\right) = -\dfrac{1}{\sqrt2}$

$\cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$

$\cos\left(\dfrac{\pi}{2}\right) = 0$

$\sin\left(\dfrac{5\pi}{4}\right) = -\dfrac{1}{\sqrt2}$

$\sin\left(\dfrac{\pi}{3}\right) = -\dfrac{\sqrt3}{2}$

$\cos\left(\dfrac{\pi}{2}\right) = 1$

Can you express it in exact form now? (Smile)