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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?

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?

Complex numbers.

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?

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

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

(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.

sebbie

how?

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

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