Welcome to NRICH.

 
Maclaurin series for sin2x
By Anonymous on Tuesday, November 28, 2000 - 06:47 pm:

How do you get the Maclaurin Series for
f(x) = sin2x

I differentiated several times, but got very messy.

By William Astle (Wja24) on Tuesday, November 28, 2000 - 07:00 pm:

The identity 2cos(x)sin(x)=sin2x will probably help you.

By Barkley Bellinger (Bb246) on Tuesday, November 28, 2000 - 07:08 pm:

Once you get to sin2x things should not start to get any messier, since differentiating from then on just results in multiples (for which you should spot a pattern) of sin2x and cos2x.

By Michael Doré (Md285) on Tuesday, November 28, 2000 - 07:21 pm:

I'm probably being a bit slow but I can't figure out how you can use 2cos(x)sin(x)=sin 2x.

You can certainly use cos2x = 1 - 2sin2x, thus:

sin2x = ½ - ½cos2x

Once you've got that you needn't even bother differentiating - just use the Maclaurin series for cos x, replace x with 2x then multiply by -½ and add ½.

By Barkley Bellinger (Bb246) on Tuesday, November 28, 2000 - 07:47 pm:

The result 2 cosx sinx = sin 2x is in connection with the first derivative of sin2x, which you need to consider if you work out the series from scratch.

By Anonymous on Tuesday, November 28, 2000 - 07:49 pm:

Thanks.
I especially like the 0.5 - 0.5cos2x sub.

By Anonymous on Tuesday, November 28, 2000 - 07:50 pm:

Thanks Barkley & William.

By William Astle (Wja24) on Tuesday, November 28, 2000 - 09:38 pm:

I think you can use sin2x = 2cos(x)sin(x) because if f(x) = sin2x then f'(x) = 2sin(x)cos(x) by the product or chain rule. However, I could be entirely wrong I never was particularly good at that sort of thing and the identity sin2x = ½ - ½cos2x works just as well. That both are as good as each other follows since d(½-½cos(2x))/dx = sin(2x)

By William Astle (Wja24) on Tuesday, November 28, 2000 - 09:40 pm:

On second thoughts Michael's answer is clearly better since, as he says you get the Maclaurin series with out doing any differentiation yourself. (Providing you know the series for cos(x).)

By William Astle (Wja24) on Tuesday, November 28, 2000 - 09:49 pm:

While I'm in the mood to send posts to this discussion do you know De Moivre's Identity? That is:

(cos(x) + isin(x))n = cos(nx) + isin(nx)

where i is a square root of -1?

This is really useful for getting trig identities, just be taking real or imaginary parts.

For instance I got sin(2x) = 2cos(x)sin(x) by taking the imaginary part with n=2.

You'll never need a formula booklet again if you know that and the addition formulae.

The proof follows since eix = cos(x) + i sin(x)