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CIE A Level's Paper 3 Question!

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sabbath_92:

--- Quote from: ifraha18 on May 13, 2010, 06:18:50 pm ---i. x=tan?
 so dx/d?=sec^2?
     dx=sec^2? d?


substitute it for dx in equation...the equation is

1_x^2/(1+x^2)^2 dx
whch becomes
1_tan^2?/(1+tan^2?)^2 sec^2 ?d?
then put in the identity fr 1+tan^2 ? as sec^2 ?

1-(sec^2?_1)/(sec^2?)^2 *sec^2 ? d?
then we get
2_sec^2?/sec^2? *d?
then
(2/sec^2?) _1 * d?
2cos^2? _1 d?
cos2? d?

--- End quote ---

Thank you so much lol. I can imagine it was a pain writing all that down lol.

BUT, how do you get

(2/sec^2?) _1 * d?
from
2_sec^2?/sec^2? *d?
That was the part I was confused with when I first tried it.

sabbath_92:
Nevermind I got it  :-[lol

ifraha18:
2_sec^2 ?/sec^2? * d?

re arranging...
this is
(2/sec^2?)_(sec^2?/sec^2?) *d?
so it becomes
(2/sec^2?)_1

n den solve de rest

sabbath_92:
These are the only identities that are needed for a2 right? Am I missing something?

tan(x y) = (tan x tan y) / (1  tan x tan y)

sin(2x) = 2 sin x cos x

cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x)

tan(2x) = 2 tan(x) / (1 - tan^2(x))

sin^2(x) = 1/2 - 1/2 cos(2x)

cos^2(x) = 1/2 + 1/2 cos(2x)

sin x - sin y = 2 sin( (x - y)/2 ) cos( (x + y)/2 )

cos x - cos y = -2 sin( (x - y)/2 ) sin( (x + y)/2 )

ifraha18:
identities r given in the mf9 fromulae sheet

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