Mathematics: Post your doubts here!

[Nouman Shafique]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
question 5 part (i) please its the 3RD TIME I AM POSTING THIS SOMEONE PLEASE HELP

cos4@ - 4cos2@
2cos^2 2@ -1 - 4cos2@
2(cos2@)^2 - 1 - 4(cos2@)
2(2cos^2 @ -1)^2 - 1 - 4(2cos^2 @ - 1)
solve it ahead u wud get the anwer, i have done the tricky part

 

[angelicsuccubus]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
question 5 part (i) please its the 3RD TIME I AM POSTING THIS SOMEONE PLEASE HELP

This is ridiculous.. two post above yours is where I answered this for the second time.. This is the third paper they've had this question in ..
here: https://www.dropbox.com/s/26xt3wmgc9jp3zj/IMAG0276.jpg

 

[hm12]

This is ridiculous.. two post above yours is where I answered this for the second time.. This is the third paper they've had this question in ..
here: https://www.dropbox.com/s/26xt3wmgc9jp3zj/IMAG0276.jpg

i opened your link thinking the same but the one i am asking is different because in this you have to prove it equal to sin^4 theeta not cos^4 theeta

 

[Mobeen]

no matter how much i do vectors still cant get any answer right ! UGH!

 

[hm12]

cos4@ - 4cos2@
2cos^2 2@ -1 - 4cos2@
2(cos2@)^2 - 1 - 4(cos2@)
2(2cos^2 @ -1)^2 - 1 - 4(2cos^2 @ - 1)
solve it ahead u wud get the anwer, i have done the tricky part

i have already done till here but still when i solve it further the answer is not coming can you please show all steps? i will be really appreciate your help

 

[angelicsuccubus]

i opened your link thinking the same but the one i am asking is different because in this you have to prove it equal to sin^4 theeta not cos^4 theeta

oh sorry.. Im really tired so I didn't see that but any way it's the same thing cause cos2A=1-2sin^2A and cos4A=1-2sin^(2)(2A) .. and you follow the same method kay?

Now Imma go sleep for a couple hours.. if you get stuck on something like this again .. try www.wolframalpha.com

 

[angelicsuccubus]

no matter how much i do vectors still cant get any answer right ! UGH!

forget it and go sleep? I can't even think straight anymore plus I suck at p3 =[ .. maybe you could find somethin on youtube?

 

[Mobeen]

forget it and go sleep? I can't even think straight anymore plus I suck at p3 =[ .. maybe you could find somethin on youtube?

other than vectors im pretty much good .. so yeah forget it ! ...
found some funny videos on youtube . or were you talking about vectors

 

[hm12]

cos4@ - 4cos2@
2cos^2 2@ -1 - 4cos2@
2(cos2@)^2 - 1 - 4(cos2@)
2(2cos^2 @ -1)^2 - 1 - 4(2cos^2 @ - 1)
solve it ahead u wud get the anwer, i have done the tricky part

okay i finally managed it was just a plus minus mistake but this question drove me mad i mean its working is too long :/

 

[angelicsuccubus]

okay i finally managed it was just a plus minus mistake but this question drove me mad i mean its working is too long :/

I just answered it.. its not that long if you switched everything to sin at the beginning.. and I suck at trig - don't forget they give you practically everything in the data booklet.

 

[angelicsuccubus]

other than vectors im pretty much good .. so yeah forget it ! ...
found some funny videos on youtube . or were you talking about vectors

lol i meant vectors.. but before you sleep.. can you please answer q7b and 8i here?
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf
please and thankyou

 

[Mobeen]

lol i meant vectors.. but before you sleep.. can you please answer q7b and 8i here?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf
please and thankyou

dont know the argand diagrams sorry !
and 8i is just binomial of the partial fraction..
partial fraction is = -2/(1+x) + (x+4)/(2+x^2)
it would be:
-2(1+x)^-1 + (x+4)(2)^-1(1+(x^2)/2)
binomial of the first fraction is easy . and binomial of the second is easy too and multiply it by (x+4)(2^-1)

 

[RGBM211]

x=sin4tetha +2sin2tetha and y=cos4tetha-2cos2tetha

where -pie/6<tetha<pie/6 show that dy/dx= -tantetha

can someone solve it

 

[hassam]

well i didnot understand the validity of binomial expansion part....cn smbody give ma quick review of it

 

[aliya_zad]

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s10_qp_31.pdf
Question 9 part i) Urgent!!Please help!!

 

[Aarjit]

so basically what I had in my head was complete insanity? .. why was I thinkin of ln?

Also, why are we "supposing" that x=tan theta? (I get that this is some ancient dervied thing.. but could x have been supposed as anything?)

So that we can cancel out sec² Ө even before we begin integrating!

1 + tan² Ө = sec² Ө ; ∫1/(1 + x²) dx ; dx = sec² Ө . dӨ............... [do you see it now? ]


∫1/(1 + x²) dx

suppose x = tan Ө so Ө = tan^(-1) x

so 1 + x² = 1 + tan² Ө = sec² Ө

and dx/dӨ = sec² Ө

dx = sec² Ө . dӨ

So ∫1/(1 + x²) dx = ∫(1/sec² Ө) . sec² Ө dӨ

= ∫ 1 dӨ

= Ө + k

= tan^(-1) x + k

 

[Mustehssun Iqbal]

Assalamu alaikum,
How to calculate argument of a complex number when it turns out to be negative?? In Q.7 part two of this past paper, the answer to argument calculated comes negative, but the answer is given in positive:
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_33.pdf

 

[zarasattar]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_31.pdf
10 b!! please help me out with this! it's urgent!

 

[smzimran]

miss irfan:

 

[angelicsuccubus]

So that we can cancel out sec² Ө even before we begin integrating!

1 + tan² Ө = sec² Ө ; ∫1/(1 + x²) dx ; dx = sec² Ө . dӨ............... [do you see it now? ]


∫1/(1 + x²) dx

suppose x = tan Ө so Ө = tan^(-1) x

so 1 + x² = 1 + tan² Ө = sec² Ө

and dx/dӨ = sec² Ө

dx = sec² Ө . dӨ

So ∫1/(1 + x²) dx = ∫(1/sec² Ө) . sec² Ө dӨ

= ∫ 1 dӨ

= Ө + k

= tan^(-1) x + k

maybe lol ..