Expansion of maclaurin series of cos^2x?

1 Answer
Apr 26, 2018

#cos^2x = 1 - x^2+ 1/3x^4 + ... =((-1)^n(2x)^(2n))/(2(2n)!)#

Explanation:

Recall that

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

Let #x = cos^2x# and #y = cos(2x)#. Then

#y = 2x -1#

#(y + 1)/2 = x#

#(cos(2x)+ 1)/2 = cos^2x#

We can now use the fact that

# cosx = 1 - x^2/(2!) + x^4/(4!) +... =sum_(n = 0)^oo(-1)^nx^(2n)/((2n)!)#

To see that

#cos(2x) = 1 - (2x)^2/(2!) + (2x)^4/(4!) + ... = sum_(n = 0)^oo (-1)^n(2x)^(2n)/((2n!)#

#cos(2x) + 1 = 2 -(4x^2)/(2!) + (2x)^4/(4!) + ... = sum_(n = 0)^oo ((-1)^n(2x)^(2n))/((2n)!)#

#(cos(2x) + 1)/2 = 1 - x^2+ 1/3x^4 + ... =((-1)^n(2x)^(2n))/(2(2n)!)#

Hopefully this helps!