How do you use the double angle or half angle formulas to derive cos(4x) in terms of cos x?

2 Answers
Sep 29, 2015

#y^' = -16sin(x)cos^3(x) +16sin^3(x)cos(x)#

Explanation:

Knowing that
#cos(2u) = cos^2(u) - sin^2(u) = 1 - 2sin^2(u)#
#sin(2u) = 2sin(u)cos(u)#

So,

#cos(4x) = 1 - 2sin^2(2x)#
#1 - 2sin^2(2x) = 1 - 2*(2sin(x)cos(x))^2#

So #cos(4x) = 1 - 4sin^2(x)cos^2(x)#

We know that the constant is irrelevant for derivatives so we can say that for the function #y = f(x)#, we have

#y = -4sin^2(x)cos^2(x)#

So then you can differentiate in any way you want, using logarithms (after taking that minus four out of the way, remembering to tack it on later), we have

#ln(y) = 2ln(sin(x)) + 2ln(cos(x))#
#y^'/y = 2cos(x)/sin(x) + 2(-sin(x))/cos(x) #
#y^' = 4sin(x)cos^3(x) -4sin^3(x)cos(x)#

Remembering to put the -4 back in

#y^' = -16sin(x)cos^3(x) +16sin^3(x)cos(x)#

Refer to explanation

Explanation:

Well it is

#cos(4x) = 2cos^2(2x) - 1 = 2[2cos^2(x) - 1]^2 - 1 = 8cos^4(x) - 8cos^2(x) + 1#