# solving a trig problem

• Aug 20th 2013, 10:36 PM
icelated
solving a trig problem

\$\displaystyle cos(7x)cos(2x) + sin(7x)sin(2x) = 1/2\$

The only place where i know where to begin is

\$\displaystyle cos(2x) = 2cos^2 - 1\$

would i need to reduce the cos(7x) like

\$\displaystyle cos(7x) = cos(4x + 3x)\$
until i get down to cos(2x)?
so i could use the identity

Can anyone help me get started on this beast?
• Aug 20th 2013, 10:43 PM
ibdutt
Re: solving a trig problem
Remember the identity cos ( A-B) = cos A cos B + sin A sin B use it and we have cos 5x = 1/2 now you can complete the solution.
• Aug 20th 2013, 10:50 PM
icelated
Re: solving a trig problem
would i then use cos(5x) = cos(3x + 2x) is that the way to go?

or would i use arccos?
5x = arccos(1/2)
• Aug 21st 2013, 12:00 AM
ibdutt
Re: solving a trig problem
No as mentioned the identity is cos ( A-B) = cos A cos B + sin A sin B
What we have is
LHS = cos(7x)cos(2x) + sin(7x)sin(2x) = cos ( 7x - 2x ) = cos 5X
thus we get cos 5x = 1/2