# trigonometry

• Apr 19th 2009, 04:16 PM
bahram
trigonometry
Hi guys
• Apr 19th 2009, 04:51 PM
mollymcf2009
Quote:

Originally Posted by bahram
Hi guys

Just to clarify the problem, is this correct?

$\displaystyle cos(3a) - \frac{cos(7a)}{sin(7a)} + sin(3a) = tan (2a)$
• Apr 19th 2009, 05:02 PM
Coren
is the question $\displaystyle \frac{{\cos3a}-{\cos7a}}{{sin7a}+{sin3a}} = {\tan2a}$

or

$\displaystyle {\cos3a}-{\frac{cos7a}{\sin7a}}+{\sin3a} = {\tan2a}$?

Your syntax is a little unclear.

Either way you may find this indentity useful: $\displaystyle {\sin3a} = {\sin (a + 2a)}$ this also applies for cos and tan.
• Apr 19th 2009, 05:15 PM
skeeter
Quote:

Originally Posted by bahram
Hi guys

$\displaystyle \frac{\cos(3a) - \cos(7a)}{\sin(7a) + \sin(3a)} =$

$\displaystyle \frac{\cos(5a - 2a) - \cos(5a + 2a)}{\sin(5a + 2a) + \sin(5a - 2a)} =$

$\displaystyle \frac{\cos(5a)\cos(2a) + \sin(5a)\sin(2a) - [\cos(5a)\cos(2a) - \sin(5a)\sin(2a)]}{\sin(5a)\cos(2a) + \cos(5a)\sin(2a) + \sin(5a)\cos(2a) - \cos(5a)\sin(2a)} =$

$\displaystyle \frac{2\sin(5a)\sin(2a)}{2\sin(5a)\cos(2a)} =$

$\displaystyle \frac{\sin(2a)}{\cos(2a)} = \tan(2a)$
• Apr 19th 2009, 05:21 PM
bahram
hi
thanks skeeter. it's the best one