I am having a hard time, how do I rewrite 3 cos 4x in terms of cos x?

Question

9674 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-26T03:12:09

In terms of #cosx#, the expression is equivalent to #24cos^4x-24cos^2x+3#.

We'll need to use this trigonometric identity, called the "cosine double angle formula":

#cos(2theta)=2cos^2theta-1#

Now here's our expression:

#color(white)=3cos(4x)#

#=3(cos(2*2x))#

#=3(2cos^2(2x)-1)#

#=6cos^2(2x)-3#

#=6(cos(2x))^2-3#

#=6(2cos^2x-1)^2-3#

#=6(2cos^2x-1)(2cos^2x-1)-3#

#=6(4cos^4x-2cos^2x-2cos^2x+1)-3#

#=6(4cos^4x-4cos^2x+1)-3#

#=24cos^4x-24cos^2x+6-3#

#=24cos^4x-24cos^2x+3#

It looks kind of weird, but you can check that they are equivalent by graphing them both and seeing that they have the same graph:

(I had to write #(cos^2x)^2# instead of #cos^4x# because Desmos doesn't understand #cos^4x#.)

That's it. Hope this helped!