# Today P2 Identity question

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#2

(Original post by

How did you do it - for those who want to try it is

cot2x+cosec2x = cotx

**lexazver203**)How did you do it - for those who want to try it is

cot2x+cosec2x = cotx

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#3

**lexazver203**)

How did you do it - for those who want to try it is

cot2x+cosec2x = cotx

so i had (cos2x+1)/sin2x=2cos^2x/2sinxcosx

=cotx

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#4

**lexazver203**)

How did you do it - for those who want to try it is

cot2x+cosec2x = cotx

=cos2x/sin2x + 1/sin2x

=(cos2x + 1)/sin2x

=(cos^2x - sin^2x + 1)/sin2x

=(cos^2x +(1 - sin^2x))/sin2x

=(2cos^2x)/2cosxssinx

=cosx/sinx

=cotx

MB

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#5

**lexazver203**)

How did you do it - for those who want to try it is

cot2x+cosec2x = cotx

(Cos2x/sin2x) + 1/(sin2x)=CotX

[(Cos^2X-Sin^2x)+(1/Sin2x)] = Cotx

cos^2X-Sin^2x + ( Sin^2x + Cos^2x)=cot X

(2cos^2X/Sin2X)=cot X

(2cox^2 X/ 2sinXcoxX)=cotx

The 2's are eliminated, Cos^2X and CosX to leave (

(CosX/SinX) which equals CotX

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ok then how about the answer for the bionomial expansion? what was the term with no x in i got it to be 495/512 and the question was

(x^3-1/2x)^12

(x^3-1/2x)^12

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#7

(Original post by

ok then how about the answer for the bionomial expansion? what was the term with no x in i got it to be 495/512 and the question was

(x^3-1/2x)^12

**lexazver203**)ok then how about the answer for the bionomial expansion? what was the term with no x in i got it to be 495/512 and the question was

(x^3-1/2x)^12

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(Original post by

no it was -55/128 but i put +55/128 so that is -1 mark (if it gets marked)

**lgs98jonee**)no it was -55/128 but i put +55/128 so that is -1 mark (if it gets marked)

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#9

(Original post by

1/tan2x + 1/sin2x

=cos2x/sin2x + 1/sin2x

=(cos2x + 1)/sin2x

=(cos^2x - sin^2x + 1)/sin2x

=(cos^2x +(1 - sin^2x))/sin2x

=(2cos^2x)/2cosxssinx

=cosx/sinx

=cotx

MB

**musicboy**)1/tan2x + 1/sin2x

=cos2x/sin2x + 1/sin2x

=(cos2x + 1)/sin2x

=(cos^2x - sin^2x + 1)/sin2x

=(cos^2x +(1 - sin^2x))/sin2x

=(2cos^2x)/2cosxssinx

=cosx/sinx

=cotx

MB

1/2 cot x + 1/2 sin x + 1/2 cot x - 1/2 sin x

that cancelled out

anyone else do it that way? (your way is nicer MB!)

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#10

**lgs98jonee**)

no it was -55/128 but i put +55/128 so that is -1 mark (if it gets marked)

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#11

YAY!!! I got 55/128!!!! Can't remember about the -ve sign - couldn't do the trig proof though

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(Original post by

OMG! I think I got something like -55/128! And I thought I'd done that whole question wrong! You have just made me sooo happy!!!!!

**Hoofbeat**)OMG! I think I got something like -55/128! And I thought I'd done that whole question wrong! You have just made me sooo happy!!!!!

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#13

hey guys, was ques 5 the binomials right with [x^3-1/2x]^12? and then 6 was that iteration stuff (zzz) and 7 was that trap rule stuff?

you know for the iteration, how did u show it was 1.70 or whatever, did u just choose 1.65 and 1.75, stick it in and show sign change ya/?

you know for the iteration, how did u show it was 1.70 or whatever, did u just choose 1.65 and 1.75, stick it in and show sign change ya/?

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#14

I just kept on expanding until i reached a part where the x's would be gone....

I prob had to use sum other method tho, dont really know

I prob had to use sum other method tho, dont really know

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#15

My method is a lot more fun. Just to awkward I used the RHS - makes it more interesting for the examiner anyway.

cotx = cosx/sinx

= 2(cosx)^2 / 2sinxcosx

From the double angle formula:

cos2x = 1 - 2(cosx)^2

2(cosx)^2 = 1 - cos2x

cotx = 1- cos2x / 2sinxcosx

= 1/2sinxcosx -cos2x/2sinxcosx

= 1/sin2x - cos2x/sin2x

= cosec2x - cot2x

cotx = cosx/sinx

= 2(cosx)^2 / 2sinxcosx

From the double angle formula:

cos2x = 1 - 2(cosx)^2

2(cosx)^2 = 1 - cos2x

cotx = 1- cos2x / 2sinxcosx

= 1/2sinxcosx -cos2x/2sinxcosx

= 1/sin2x - cos2x/sin2x

= cosec2x - cot2x

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(Original post by

I just kept on expanding until i reached a part where the x's would be gone....

I prob had to use sum other method tho, dont really know

**toxi**)I just kept on expanding until i reached a part where the x's would be gone....

I prob had to use sum other method tho, dont really know

so then x will cancel at

x^36(12*11*10*9*8*7*6*5/8! * 1/((2^9)(x^36)) and it cancelled down to 495/512 and it is -ve

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#18

(Original post by

Can anyone show me what the graph of ln|3x-6| looks like?

**imasillynarb**)Can anyone show me what the graph of ln|3x-6| looks like?

Asymptote at x = 2

line coming up from the asymptote on the right crossing the x axis at 7/3 (i think) then the curve slowed down but y continued to increase with x

then this was sort of reflected in the asymptote crossing x axis at 5/3 and y axis at ln 6

rosie

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(Original post by

If you can imagine it this is how I drew it:

Asymptote at x = 2

line coming up from the asymptote on the right crossing the x axis at 7/3 (i think) then the curve slowed down but y continued to increase with x

then this was sort of reflected in the asymptote crossing x axis at 5/3 and y axis at ln 6

rosie

**crana**)If you can imagine it this is how I drew it:

Asymptote at x = 2

line coming up from the asymptote on the right crossing the x axis at 7/3 (i think) then the curve slowed down but y continued to increase with x

then this was sort of reflected in the asymptote crossing x axis at 5/3 and y axis at ln 6

rosie

i thought it would look like that

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#20

the easisest way to do the question was tht x was decreasing by

X^(36-4x)

when 36-4x = 0

x=9

hence it was the 10th term

since its the tenth term use 12C9 * (-1/512)

X^(36-4x)

when 36-4x = 0

x=9

hence it was the 10th term

since its the tenth term use 12C9 * (-1/512)

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