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# sin5x equal to... watch

1. a) 5sinx-20sin^3x+16sin^5x

b) 5sinx-20sin^3x+14sin^5x

c) 5sinx-10sin^2x+10sin^3x-5sin^4x+sin^5x

d) sinx-5sin^2x+10sin^3x-10sin^4x+5sin^5x

The answer is a) but the question is on an old Oxford exam.... Most of the questions for the same number of marks didn't take anywhere near as long to work out, the method I used for this was quite laborious... Is there an easier way?
2. Complex numbers.
3. how?
4. (Original post by sebbie)
a) 5sinx-20sin^3x+16sin^5x

b) 5sinx-20sin^3x+14sin^5x

c) 5sinx-10sin^2x+10sin^3x-5sin^4x+sin^5x

d) sinx-5sin^2x+10sin^3x-10sin^4x+5sin^5x

The answer is a) but the question is on an old Oxford exam.... Most of the questions for the same number of marks didn't take anywhere near as long to work out, the method I used for this was quite laborious... Is there an easier way?
it's fairly easy to work out which is correct by a process of elimination

(c) and (d) aren't odd but the given function is

(b) disagrees with the given function at pi/2 giving 5-20+14 = -1 rather than 1
5. (Original post by RichE)
it's fairly easy to work out which is correct by a process of elimination

(c) and (d) aren't odd but the given function is

(b) disagrees with the given function at pi/2 giving 5-20+14 = -1 rather than 1
Aaah, that's much more clever.
6. (Original post by sebbie)
how?
You may know that . Use the binomial expansion on this and mess around a bit, you should end up with . You can rewrite the even powers of cos in terms of sine using the Pythagorean identity (note there will only be even powers of sine in the expansion of an odd n, you should see why from the first sine term in the last formula I wrote). You can evaluate this expression for n=5.

Anyway, RichE's method is of course a much better approach (and the kind of approach I wouldn't think of )

In other news, I officially despise mimetex
7. \huge

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