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Syncman
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#1
Report Thread starter 14 years ago
#1
Please help. Im stuck its been a long time since i did reduction formula and im trying to revise it. But im stuck on:

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BCHL85
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#2
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Ok, use this :sinnx = sin[(n-1)x +x] = sin[(n-1)x]cosx + cos[(n-1)x]sinx
and continue with cos(n-1)x = cos[(n-2)x +x] = cos[(n-2)x]cosx - sinxsin[(n-2)x]
That gives you the answer
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Syncman
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ok i just tried that and now im in more of a mess that before gawd i hate reduction formula!

So to get it straight. I sub in the above then intergrate using parts?
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Jonny W
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#4
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Using sin(A) - sin(B) = 2 cos[(1/2)(A + B)] sin[(1/2)(A - B)],

I(n) - I(n - 2)
= (int from 0 to 1) [sin(nx) - sin((n - 2)x)] / sin(x) dx
= (int from 0 to 1) 2 cos((n - 1)x) sin(x) / sin(x) dx
= (int from 0 to 1) 2 cos((n - 1)x) dx
= [ 2 sin((n - 1)x) / (n - 1) ] (from 0 to 1)
= 2 sin(n - 1) / (n - 1)
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Bilal
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How come everyone here are geniuses at Maths? :confused: I feel so dumb.
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Syncman
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#6
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lol dont worry we can be dumb together i couldnt do it either
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