Question

What is the Maclaurin series for `cos(sinx)`?

Answer 1

`cos(sinx) = 1 -1/2x^2 + ...`

Explanation:

Let:

`f(x) = cos(sinx)` ..... [A]

The Maclaurin series is given by

`f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...`

First Term:

`f(0) = cos(sin0)=cos0=1`

Second Term:

Differentiating [A] wrt `x`
`f'(x) = -cos(x)*sin(sin(x))` ..... [B]
`:. f'(0) = cos0 sin(sin0) =0`

Differentiating [B] wrt `x`:

`f''(x) = sin(x)sin(sin(x))-cos(x)^2cos(sin(x))`
`:. f''(0) = sin(0)sin(sin(0))-cos(0)^2cos(sin(0)) = -1`

.... etc for higher derivatives

So, the power series is given by
`f(x) = f(0) + (f'(0))/(1!) + (f''(0))/(2!) + (f'''(0))/(3!) + ...`

`:. f(x) = 1 + (0)/(1)x + (-1)/(2)x^2 + ...`
`\ \ \ \ \ \ \ \ \ \ \ \ = 1 -1/2x^2 + ...`