Question

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

Answer 1

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

`color(red)or`

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

Explanation:

I am assuming that `cos(x)^2` refers to `cos^2(x)`.

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First, find the first 4 or 5 derivatives of `f(x) = cos^2(x)`

This is much easier if we use the identity `cos^2(theta) = (1+cos(2theta))/2`.

Therefore, the first few derivatives of `f(x)` are:

`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 `0` to each derivative.

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