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**Tweety** $\displaystyle sin^{3}x = \frac{3}{4} sinx - \frac{1}{4}sin(3x) $ Use this identity and the Maclaurin series for sin x to

show that

$\displaystyle sin^{3} x = x^{3} - \frac{1}{2}x^{5} + \frac{13}{120}x^{7} $

I am quite stuck,

I know that sinx = $\displaystyle x-\frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} $

and I know sin3x = sin(2x+x)

so I have

$\displaystyle \frac{3}{4}( x-\frac{x^{3}}{3!} + \frac{x^{5}}{5!}) - \frac{1}{4} (sin2x +x) $

however when I expand sin2x +x I just it just gets more complicated, is this right so far? what should I do next ?

Thank you.

We have the identity

sin^3(x) = 3/4sin(x)-1/4sin(3x)

And want to use Maclaurin expansion to show that

sin^3(x) = x^3-1/2x^5+13/120x^7

Since we know that the expansion of sin(x) is

sin(x) = x-x^3/3!+x^5/5!-x^7/7! **(1)**

We can raise both sides to the power or three hence we have

sin^3(x) = (x-x^3/3!+x^5/5!-x^7/7!)^3

But since we also know that

sin^3(x) = 3/4sin(x)-1/4sin(3x) **(2)**

By using the relationship **(1)** into **(2)** we get

sin^3(x) = 3/4*(x-x^3/3!+x^5/5!-x^7/7!)-1/4*((3x)-(3x)^3/3!+(3x)^5/5!-(3x)^7/7!) = x^3-1/2x^5+13/120x^7