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Mathematics Paper 3: Post your doubts!

Amy Bloom

Amy Bloom

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Mattman said:
eventually u get tan()=tan()
tan^-1(tan()) is theta.

when they mean constant, it means that it is a real number, and not sintheta or something, where as the angle changes, the real part changes.

Just expand it normally and ull find the awnser.


For the partial question question, youy notice there is an extra "x" in the equation. This means the LHS numerator needs to be multiplied by X as well, which is(2x^3-1)(x)
Click to expand...
Hey thanks buddy ;) but one thing i don't get. can you show me a few lines of your workings for what i have put in green font
 
M

Mattman

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Amy Bloom said:
Hey thanks buddy ;) but one thing i don't get. can you show me a few lines of your workings for what i have put in green font
Click to expand...
Its 1/z, a fraction, so u multiply it with a conjugate, which is (1+cos2t) MINUS isin2t, which is z*/z(z*)

since were only interested in the real part, you do (1+cos2t)/[(1+cos2t)+isin2t][(1+cos2t) - isin2t], apply trigo identities here, sin^2(2t)+cos^2(2t)=1

in the end youll get (1+cos2t)/2+2cos2t, which by simpliflying, get 1/2
 
Amy Bloom

Amy Bloom

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Mattman said:
Its 1/z, a fraction, so u multiply it with a conjugate, which is (1+cos2t) MINUS isin2t, which is z*/z(z*)

since were only interested in the real part, you do (1+cos2t)/[(1+cos2t)+isin2t][(1+cos2t) - isin2t], apply trigo identities here, sin^2(2t)+cos^2(2t)=1

in the end youll get (1+cos2t)/2+2cos2t, which by simpliflying, get 1/2
Click to expand...
:cool: Cool! you're a boss!
Thanks
 
M

Mattman

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its ln(x)/sqrt(x), so its just x^-.5ln(x), you move the sqrt(x) up basically.

its like x/y becomes xy^-1
 
Amy Bloom

Amy Bloom

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Mattman said:
its ln(x)/sqrt(x), so its just x^-.5ln(x), you move the sqrt(x) up basically.

its like x/y becomes xy^-1
Click to expand...
Alright. You do the following right? Confirm this for me plz... i'm in trouble.
U = lnx dv/dx = x^-0.5
du/dx = 1/x V=2sqrtx
 
AhShun

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