Re: Maclaurin series help

Quote:

Originally Posted by

**Tweety** Use this identity and the Maclaurin series for sin x to

show that

I am quite stuck,

I know that sinx =

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

so I have

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

Re: Maclaurin series help