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

Question

7979 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-09T01:50:21

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

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 + ... #