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

Question

Cameron Alexander · Accepted Answer

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

Let:

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

The source of the Maclaurin series is

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

Initial Term:

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

Term Two:

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 #

For higher derivatives, see.... etc.

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 + ... #

Ezra Adams · Answer

undefined The Maclaurin series expansion for $ \cos(\sin(x)) $ can be found by substituting $ \sin(x) $ into the Maclaurin series for $ \cos(x) $:

$$ \cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} $$

Substituting $ \sin(x) $ for $ x $, we get:

$$ \cos(\sin(x)) = \sum_{n=0}^{\infty} (-1)^n \frac{\sin(x)^{2n}}{(2n)!} $$

However, we can express $ \sin(x) $ in terms of its Maclaurin series to get a more concise representation:

$$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} $$

Thus, the Maclaurin series for $ \cos(\sin(x)) $ becomes:

$$ \boxed{\cos(\sin(x)) = \sum_{n=0}^{\infty} (-1)^n \frac{\left(\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}\right)^{2n}}{(2n)!}} $$

This expression represents the Maclaurin series expansion for $ \cos(\sin(x)) $.