# Find the Maclaurin series for f(x) = cos(x^{3}) and use it to determine f(6).

**Solution:**

Given f(x) = cos(x^{3})

We know that the Maclaurin series for f(x) = f(0)+f'(0)x+f''(0)/2!x2+f'''(0)/3!+f''''(0)/4!+⋯

\(\cos (x)=1-\dfrac{x^{2}}{2 !}+\dfrac{x^{4}}{4 !}-\dfrac{x^{6}}{6 !}+\cdots=\sum_{n=0}^{\infty} \dfrac{(-1)^{n} x^{2 n}}{(2 n) !}\)

Let x^{3} = p

We know cosp = 1- x^{2}/2! +x^{4}/4! - x^{6}/6!

cos(x³) = 1- (x³)²/2! + (x³)4/4! - (x³)⁶/6!

cos(x³) = 1- x⁶/2! + x¹²/4! - x¹⁸/6!

= 1- ½ x⁶ +⋯

x=6 in f(x) doesn’t affect the maclaurin series as it is defined only at origin.

## Find the Maclaurin series for f(x) = cos(x^{3}) and use it to determine f(6).

**Summary :**

The Maclaurin series for f(x) = cos(x3) = 1- ½ x⁶ +⋯