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Mathematics: Post your doubts here!

Banglarbagh

Banglarbagh

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Hello can anyone solve this maths for me.... June 12/p32 Question no. 2 (ii) .... Please help me......
 
Maz

Maz

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Banglarbagh said:
Hello can anyone solve this maths for me.... June 12/p32 Question no. 2 (ii) .... Please help me......
Click to expand...
Take f(x) = ø - tan^-1(3ø)

Search for a sign change.
f(0) = 0
f(pi/4) = -0.38
f(pi/2) = 0.21

Therefore, the root is between pi/4 and pi/2...
take x1 = pi/4...
I guess now it's easy...
 
Banglarbagh

Banglarbagh

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Maz said:
Take f(x) = ø - tan^-1(3ø)

Search for a sign change.
f(0) = 0
f(pi/4) = -0.38
f(pi/2) = 0.21

Therefore, the root is between pi/4 and pi/2...
take x1 = pi/4...
I guess now it's easy...
Click to expand...


Brother i still did not get u..... how do i do it.... at first i take f(O) = O .... den what ? :/
 
kaushar

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am having a prob with paper 13 may/june 2013 i cant understand where it got ½.3²π have it in the marking scheme plzz help number 2
 
$~SauD~$

$~SauD~$

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Banglarbagh said:
Brother i still did not get u..... how do i do it.... at first i take f(O) = O .... den what ? :/
Click to expand...

dude dude.. It's simple..

Equation is θn+1 = tan^−1 (3θn) between 0< θ <.5π
now take any value between that range... let's take .25π

θ1 = .25π
θ2 = tan^−1 (3(.25π)) = 1.1694
θ3 = tan^-1 (3(1.1694)) = 1.2931 << is the 'ans' right?
θ4 = tan^-1 (3(ans)) = 1.3185
θ5 = tan^-1 (3(ans)) = 1.3231
θ6 = tan^-1 (3(ans)) = 1.3240 do it until we get all 4 decimal point same :)
θ7 = tan^-1 (3(ans)) = 1.3241
θ8 = tan^-1 (3(ans)) = 1.3241 <<<< this and above have same.. so the answer should be written in 2 decimal as said by question

so the final answer is 1.32
 
Banglarbagh

Banglarbagh

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$~SauD~$ said:
dude dude.. It's simple..

Equation is θn+1 = tan^−1 (3θn) between 0< θ <.5π
now take any value between that range... let's take .25π

θ1 = .25π
θ2 = tan^−1 (3(.25π)) = 1.1694
θ3 = tan^-1 (3(1.1694)) = 1.2931 << is the 'ans' right?
θ4 = tan^-1 (3(ans)) = 1.3185
θ5 = tan^-1 (3(ans)) = 1.3231
θ6 = tan^-1 (3(ans)) = 1.3240 do it until we get all 4 decimal point same :)
θ7 = tan^-1 (3(ans)) = 1.3241
θ8 = tan^-1 (3(ans)) = 1.3241 <<<< this and above have same.. so the answer should be written in 2 decimal as said by question

so the final answer is 1.32
Click to expand...

Thanks man i got it fully! :D
 
Kumkum

Kumkum

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Banglarbagh said:
Can some one help me with june 12 p32/ No. 4..... please.....
Click to expand...

cosec(2x) = secx + cotx
1/sin(2x) = 1/cosx + cosx/sinx [using the identity; sin(2x) = 2sinxcosx on the right hand side and taking the common denominator on the left hand side]
===> 1/2sinxcosx = (sinx + (cos^2)x)/cosxsinx [sinxcosx cancels out]
1 = 2sinx + 2(cos^2)x [using identity (sin^2)x + (cos^2)x = 1 on left hand side]
1 = 2sinx + 2(1 - sin^2 x)
1 = 2sinx + 2 - 2sin^2 x
2sin^2 x - 2sinx - 1 =0 (i'm using the quadratic formula since there are no factors)

sinx = -(-2) +/- sq.root [(-2)^2 - ( 4*2*-1)] / 2*2
sinx = [2+/-sq.rt 12] / 4

so, sinx = [2+ sq.rt 12] / 4
sinx = 1.366 ( this is out)

or
sinx = [2 - sq.rt 12] / 4
sinx = -0.366 (i often say let sin(a) = 0.366, so that i can work with positive angles)
sin(a) = 0.366
y = 21.47 <----use this angle to find the other angles where 'sin' is negative
u'll get x as 201.5 and 338.5

hope u got it :)
 
Banglarbagh

Banglarbagh

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Kumkum said:
cosec(2x) = secx + cotx
1/sin(2x) = 1/cosx + cosx/sinx [using the identity; sin(2x) = 2sinxcosx on the right hand side and taking the common denominator on the left hand side]
===> 1/2sinxcosx = (sinx + (cos^2)x)/cosxsinx [sinxcosx cancels out]
1 = 2sinx + 2(cos^2)x [using identity (sin^2)x + (cos^2)x = 1 on left hand side]
1 = 2sinx + 2(1 - sin^2 x)
1 = 2sinx + 2 - 2sin^2 x
2sin^2 x - 2sinx - 1 =0 (i'm using the quadratic formula since there are no factors)

sinx = -(-2) +/- sq.root [(-2)^2 - ( 4*2*-1)] / 2*2
sinx = [2+/-sq.rt 12] / 4

so, sinx = [2+ sq.rt 12] / 4
sinx = 1.366 ( this is out)

or
sinx = [2 - sq.rt 12] / 4
sinx = -0.366 (i often say let sin(a) = 0.366, so that i can work with positive angles)
sin(a) = 0.366
y = 21.47 <----use this angle to find the other angles where 'sin' is negative
u'll get x as 201.5 and 338.5

hope u got it :)
Click to expand...


Thanks Sis got it...i just goofed up in the part where it shuld have been sinx + (cos^2)x ....thats why i culdnt solve it...newys thanks! :D
 
kaushar

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$~SauD~$ said:
(i) Area of circle = (pie) x r^2
Area of shaded = 1/2 x 9^2 x Q - 1/2 x 3^2 x Q

area of circle = area of shaded region

(pie) x 3^2 = .5 x81 x Q - .5 x 9 x Q

now do the equation and get Q. and Q is Theeta :D
btw, Paper 1? :s
Click to expand...

ohh thnks i just mess it up just because of stress and exams
 
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