G'day, Pingu.
f(x)=cos(2x)⋅ cosx+sin(2x)⋅sinx passes as a function, so it may better to ask to write the expression as a single trigonometric function.
By that I mean to simplify it down to just
cos(x) or just
sin2(2x), for examples, as opposed to having both sin and cos in there.
A lot of this work is a bit of trial and error, playing with identities and hoping you get a nice result out. It can be quite difficult to see the end of the path from the beginning; you can just make your way until you can see it. With practise, your foresight improves.
One option to start with would be to put
cos(2x) and
sin(2x) in terms of
cosx and
sinx, respectively:
You should be familiar with the following identities
. . \(\displaystyle \L \cos{2x} = 2\cos^2{x} - 1\)
. . \(\displaystyle \L \sin{2x} = 2\sin{x}\cos{x}\)
So
. . \(\displaystyle \L \cos{2x} \cdot \cos{x} + \sin{2x} \cdot \sin{x}\)
becomes
. . \(\displaystyle \L (2\cos^2{x} - 1)\cos{x} \, + \, (2\sin{x}\cos{x}) \cdot \sin{x}\)
Simplify:
. . \(\displaystyle \L 2\cos^3{x} \, - \, \cos{x} \, + \, 2\sin^2{x}\cos{x}\)
See if you click onto the next step. (Give it a few minutes.)