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# prooving P6 watch

1. assuming that

z^n - z^-n = 2i sin nθ

how do you proove that

16sin^5 θ = sin 5θ - 5 sin3θ + 10 sinθ
2. (Original post by lesser weevil)
assuming that

z^n - z^-n = 2i sin nθ

how do you proove that

16sin^5 θ = sin 5θ - 5 sin3θ + 10 sinθ
3. Thanks Gaz - but I don't get the 2nd to last and last lines...
4. The second to last line comes merely from expanding the brackets and grouping some of the terms differently [ie collecting your similar expressions that you need to convert back to trigonometric functions.
The last line comes from the result on the first line, taking n as 1,3 and 5 as and when appropriate.
5. oh, yeah, ok, I've got it. Thanks.
6. The last step is to notice that the three bracketed terms follow the sin(n theta) pattern - see the first line.

*EDIT* That'll teach me not to reply to an "old" page..

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