How do you find the maclaurin series expansion of #sin^3 (x)#?

Question

Isaiah Allen · Accepted Answer

#sin^3(x) = x^3 + frac{1}{2}x^5 + frac{13}{120}x^7 + ...#

This is the process of building a Maclaurin series.

#f(x) = f(0) + frac{f'(0)}{1!}x + frac{f''(0)}{2!}x^2 + frac{f^((3))(0)}{3!}x^3 + ...#

Only the first three non-zero terms will I complete.

#f(0) = sin^3 (0) =0#

#f'(x) = 3sin^2(x)cos(x)#

#f'(0) = 3sin^2(0)cos(0)#

#= 0#

#f''(0) = 0#

#f^((3))(0) = 6#

#f^((4))(0) = 0#

#f^((5))(0) = 60#

#f^((6))(0) = 0#

#f^((7))(0) = 546#

Thus, the series' continuation is

#sin^3(x) = frac{f^((3))(0)}{3!}x^3 + frac{f^((5))(0)}{5!}x^5 + frac{f^((7))(0)}{7!}x^7 + O(x^8)#

#= frac{6}{6}x^3 + frac{60}{120}x^5 + frac{546}{5040}x^7 + O(x^8)#

#= x^3 + frac{1}{2}x^5 + frac{13}{120}x^7 + O(x^8)#

Evelyn Agard · Answer

undefined To find the Maclaurin series expansion of $ \sin^3(x) $, we first use the identity $ \sin^3(x) = (\sin(x))^3 $. Then, we expand $ (\sin(x))^3 $ using the binomial theorem, which states that $ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k $.

Expanding $ (\sin(x))^3 $ using the binomial theorem with $ a = \sin(x) $ and $ b = 0 $, we get:

$ (\sin(x))^3 = \binom{3}{0}(\sin(x))^3 + \binom{3}{1}(\sin(x))^2(0) + \binom{3}{2}(\sin(x))^1(0)^2 + \binom{3}{3}(\sin(x))^0(0)^3 $

Simplify each term:

$ = \sin^3(x) + 0 + 0 + 0 $

Therefore, the Maclaurin series expansion of $ \sin^3(x) $ is simply $ \sin^3(x) $.