How do you find the Maclaurin Series for #sin^3 (x)#?

Question

Scarlett Allison · Accepted Answer

#sin^3 x = sum_(n=0)^oo (((-1)^n3(1-3^(2n)) )/(4(2n+1)!)x^(2n+1))#

#= x^3-x^5/2+(13x^7)/120-(41x^9)/3024+...#

It is probably easiest to derive it from the Maclaurin series for #sin x# and the triple angle formula for #sin#: #sin x = sum_(n=0)^oo ((-1)^n )/((2n+1)!)x^(2n+1) = x - x^3/(3!) + x^5/(5!) - x^7/(7!) +...# #sin 3 x = 3 sin x - 4 sin^3 x# So: #sin^3 x = 1/4 (3 sin x - sin 3 x)# #color(white)(sin^3 x) = 1/4 (3 (sum_(n=0)^oo ((-1)^n x^(2n+1))/((2n+1)!)) - (sum_(n=0)^oo ((-1)^n (3x)^(2n+1))/((2n+1)!)))# #color(white)(sin^3 x) = 1/4 sum_(n=0)^oo (3((-1)^n x^(2n+1))/((2n+1)!)- ((-1)^n (3x)^(2n+1))/((2n+1)!))# #color(white)(sin^3 x) = 1/4 sum_(n=0)^oo (((-1)^n(3-3^(2n+1)) x^(2n+1))/((2n+1)!))# #color(white)(sin^3 x) = sum_(n=0)^oo (((-1)^n3(1-3^(2n)) )/(4(2n+1)!)x^(2n+1))#

Axel Alvarez · Answer

undefined To find the Maclaurin series for $ \sin^3(x) $, we'll start with the Maclaurin series for $ \sin(x) $, which is $ x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $. Then, we'll cube this series term by term, and simplify to find the Maclaurin series for $ \sin^3(x) $.

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

Using the binomial theorem, we expand $ (\sin(x))^3 $ as $ \sin^3(x) = \sin(x) \cdot \sin^2(x) $.

Then, $ \sin^2(x) = (1 - \cos(2x))/2 $.

Finally, replace $ \sin(x) $ with its Maclaurin series and simplify to find the Maclaurin series for $ \sin^3(x) $.

Samantha Adkison · Answer

undefined To find the Maclaurin series for $ \sin^3(x) $, we first need to express it in terms of a known Maclaurin series. We can use the trigonometric identity $ \sin^3(x) = (\sin(x))^3 $ and then expand $ \sin(x) $ using its Maclaurin series.

The Maclaurin series for $ \sin(x) $ is:

$$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$

To find $ (\sin(x))^3 $, we raise each term in the series to the power of 3 and then multiply the resulting series term by term.

$$ (\sin(x))^3 = (x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots)^3 $$

Expanding this expression, we get the Maclaurin series for $ \sin^3(x) $.

Note that the Maclaurin series for $ \sin^3(x) $ may involve a large number of terms, as it requires expanding the cube of each term in the Maclaurin series for $ \sin(x) $ and then combining like terms.