# cos x? its supposedly simple....

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#1
lol nm i worked it out...
0
15 years ago
#2
(Original post by Dust)
how can u convert (1-2X^2)^-3

all into

1-3(-2X^2)

...it doesnt make sense?
Using the binomial ytheorem, if x is sufficiently small that terms in x^4 and higher can be ignored, (1-2x^2)^(-3) = 1 + 2x^2 (approx)
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#3
lol yup... thats wat i got in the end...
thanx

lol i always panic wen i see cos sin's etc lol
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#4
okay how can u convert the integral of x cos^2 X into 1/2 of the intergral of X cost 2X +x?

i can see how u get X cost 2X but where did the +X come from at the end?
0
15 years ago
#5
(Original post by Dust)
okay how can u convert the integral of x cos^2 X into 1/2 of the intergral of X cost 2X +x?

i can see how u get X cost 2X but where did the +X come from at the end?
cos²x = ½(1 + cos2x)
x.cos²x = ½x(1 + cos2x)
x.cos²x = ½(xcos2x + x)
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#6
lol oh yeah.. lol now i feel stupid
0
15 years ago
#7
(Original post by Dust)
lol oh yeah.. lol now i feel stupid
obviously you've just been working too hard, that's all
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#8
(Original post by Fermat)
obviously you've just been working too hard, that's all
lol i wish i could believe that.. *shrugs*

okay i dont get this either: how does this work

when -2K=ln 1/2
how does that make k= ln 2??
0
15 years ago
#9
-2k = ln(1/2) = ln1 - ln2 = -ln2
2k = ln2
k = (ln2)/2
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#10
lol ah that makes sense...
cheers
0
15 years ago
#11
Could be ln(sqrt2).
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#12
(Original post by ZJuwelH)
Could be ln(sqrt2).
how?
0
15 years ago
#13
(Original post by Dust)
how?
because,

plnx = ln(x^p)

for example,

5ln2 = ln2 + ln2 + ln2 + ln2 + ln2

but lnA + lnB = ln(A*B), (by definition of logs)
So,

ln2 + ln2 + ln2 + ln2 + ln2 = ln(2*2*2*2*2)
ln2 + ln2 + ln2 + ln2 + ln2 = ln(2^5)
5ln2 = ln(2^5)
=========

So,

½lnx = ln(x^½) = ln(√x)
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