math. =( help!
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assuming we have this trigo function put to high powers...
lets say we have (sinx)^5 ..
how are we supposed to convert this function: (sinx)^5 to single powers of sine (by manipulating the values of x)?
for example , we have: (sinx)^5 = asin(3x) + bsin(5x) + c(sin5x)....
so what i meant was.. how do we get the RHS??
i think its A lvl work.. but i forgot alrdy *sighz* dun have my jc maths notes cos i threw them away back then after i exam..
for those who can help... many many thanks.
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chim leh...
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sin^5(x)
= sin²x * sin³x
= (sin²x)(1/4)(3 sin x - sin 3x)
= 3/4 sin³x - (1/4)(sin 3x)(sin²x)
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if you are trying to integrate sin(x)^5,
you can break it into sin(x)^4*sin(x),
= (sin(x)^2)^2 * sin(x)
= (1-cos(x)^2)^2 *sin(x)
let u=cos(x), notice that du becomes -sin(x).
So you whole expression becomes
- Intergate {(1-u^2)^2 du}
= - Integare {(1-2u^2+u^4) du}
hi sorry think i am not answering your question -
maybe u can split (sin x)^5 to (sin x)^3*(sin x)^2?
i pretty sure these two terms can be converted to sin with power 1
but not sure about the product
i just finished my A-lvl maths...dun wanna think abt it
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haha thanks for ur help guys, actually i know of the reduction of sinx by using the expansion sin^3 (do until like siao?
)..
anyway i remembered the shortcut method of doing liao, its to use de mouvier's theorem (complex numbers).
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oooooo....
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man i dun understand this...
I gonna learn complex number soon and this make me scare =.= -
olvl a math?
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No.. A level O maths..Originally posted by Orange*Asterisk:olvl a math? -
Let z = cos x + isinx = c + is
z^5 = (c + is)^5
= .... using binomial expansion
= c^5 + i(5c^4s) - (10c^3s^2) - i(10c^2c^3) + 5cs^4 + is^5
Using de movire's theorem, z^5 = cos 5x + isin5x
So comparing imaginary parts, sin 5x = s^5 - 10c^2s^3 + 5c^4s
= 16s^5 - 20s^3 + 5s -
but you want to reduced sin(x)^5, not sin 5xOriginally posted by Selenite`:Let z = cos x + isinx = c + is
z^5 = (c + is)^5
= .... using binomial expansion
= c^5 + i(5c^4s) - (10c^3s^2) - i(10c^2c^3) + 5cs^4 + is^5
Using de movire's theorem, z^5 = cos 5x + isin5x
So comparing imaginary parts, sin 5x = s^5 - 10c^2s^3 + 5c^4s
= 16s^5 - 20s^3 + 5s -
Originally posted by d3sT1nY:No.. A level O maths..
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tsk tsk fmaths student leh...how can you forget de movires theorem?
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use the basics...
like Cos2x = 2cos^2(x) -1 = cos^2(x) - sin^2(x)
and many of these....
you already can see from the above that the equation has related the cos 2x to the higher power of cos^2(x).
Thus, e.g. Sin^5 (x) = sin^2(x) * sin^3(x)
= (1+cos^2(x)) * sin^2(x) * sin(x)
etc lah....
eventually you will get single power representations of Sin^5(x). -
read wrongly..
Use the results z^n - 1/z^n = 2isin nx, where z = cos x + isinx
So, let (2isinx)^5 = (z - 1/z)^5
= expand using binomial
= z^5 - 5z^3 + 10z - 10/z + 5/z^3 - 1/z^5
= (z^5 - 1/z^5) - 5(z^3 - 1/z^3) + 10(z - 1/z)
= 2i sin5x - 5(2i sin3x) + 10(2isinx)
=> 32i(sinx)^5 = 2i sin5x - 5(2i sin3x) + 10(2isinx)
(sinx)^5 = 1/16 [ sin 5x - 5sin 3x + 10sin x ] -
Originally posted by Selenite`: read wrongly..
Use the results z^n - 1/z^n = 2isin nx, where z = cos x + isinx
So, let (2isinx)^5 = (z - 1/z)^5
= expand using binomial
= z^5 - 5z^3 + 10z - 10/z + 5/z^3 - 1/z^5
= (z^5 - 1/z^5) - 5(z^3 - 1/z^3) + 10(z - 1/z)
= 2i sin5x - 5(2i sin3x) + 10(2isinx)
=> 32i(sinx)^5 = 2i sin5x - 5(2i sin3x) + 10(2isinx)
(sinx)^5 = 1/16 [ sin 5x - 5sin 3x + 10sin x ] i also read wrongly....
i think your are right!