# How do you find the Maclaurin Series for #cos (x)^2#?

##### 1 Answer

#color(red)or#

#### Explanation:

I am assuming that

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

First, find the first 4 or 5 derivatives of

This is much easier if we use the identity

Therefore, the first few derivatives of

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

#f'(x) = -sin(2x)#

#f''(x) = -2cos(2x)#

#f'''(x) = 4sin(2x)#

#f^((4))(x) = 8cos(2x)#

Since we are finding a Maclaurin series, plug in

#f(0) = (1+cos(0))/2 = 1#

#f'(0) = -sin(0) = 0#

#f''(0) = -2cos(0) = -2#

#f'''(0) = 4sin(0) = 0#

#f^((4))(0) = 8cos(0) = 8#

It is easy to see that the coefficients will continue on in this way, with odd derivatives always being 0, and even derivatives being 2 to the power of one less than the order of the derivative, with alternating positive and negative signs. Therefore, the Maclaurin series can be written as:

#cos^2(x) = 1 - (2x^2)/(2!) + (8x^4)/(4!) - (32x^6)/(6!)+...#

Or, using summation notation and a little bit of intuition,

#cos^2(x) = 1+sum_(k=1)^(oo) ((-1)^(k)(2)^(2k-1))/((2k)!) * x^(2k)#

*Final Answer*

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(You can check for yourself with a graphing interface like Desmos to see if this works!)