Need help verifying this(trig related):

januaryair · #1 $Old$ January 6th, 2009, 03:48 PM

cos(A+B)= cosAcosB - sinAsinB

for which A= (pi)/6
and B= (pi)/3

thank you, thank you, thank you

skeeter · #2 $Old$ January 6th, 2009, 03:53 PM

$\cos\left(\frac{\pi}{6} + \frac{\pi}{3}\right) = \cos\left(\frac{\pi}{2}\right) = 0$

$\cos\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{3}\right)$

look at your unit circle and evaluate the above expression ... you should get 0, right?

craig · #3 $Old$ January 6th, 2009, 03:58 PM

$a = \frac{\pi}{6}$ and $b= \frac{\pi}{3}$

$\cos( \frac{\pi}{2}) = 0$

$\cos( \frac{\pi}{6}).\cos( \frac{\pi}{3})-\sin( \frac{\pi}{6}).\sin( \frac{\pi}{3}) \rightarrow \frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

This what you mean?

craig · #4 $Old$ January 6th, 2009, 03:58 PM

Sorry I was beaten to it

januaryair · #5 $Old$ January 6th, 2009, 03:59 PM

Yep, thats right. Thank you. i had the second part equaling zero, but i couldnt get the first part. You helped a lot!