Toughest P1 Math questions I came across. A little help...?

[SZ11]

So these are the toughest P1 questions I came across and I'm having a bit of trouble solving them. A little help would be greatly appreciated.

NB: the '^' sign stands for 'raised to the power of'.

Q1)
Prove the identity.
sin^2x [sin square x] - cos^2x [cos square x] tanx - 1
/ [divided by] = /[divided by]
1 + 2sinxcosx tanx + 1




q2) the region is rotated through 360 degrees about x-axis. Calculate the volume of the shaded region.

the image : http://s941.photobucket.com/albums/ad25 ... 010025.jpg
that's the questions graph.
the image: http://s941.photobucket.com/albums/ad25 ... 010025.jpg
that's the question graph.

 

[ds29]

Re: Toughest P1 Math questions I came across. A little help.

I didn't quite understand the formula you've put for the trignometric one.

But I'll help you with the curve question.

First Volume of Revolution is (Integration) Pi Y^2
So Y^2 = 36/(x^2)
Integrate that with the limits 6 and 2.

You'd get [-36(X)^-1] to limits 6 and 2
which gives u 12pi

However there is an additional rectangular area not covered by the curve. That is represented by Y = 3
Y^2 = 9
Integrate it with limits 3 and 0 and you'd get [9X]
which gives you 27pi

Therefore total Volume = 27pi + 12pi = 39pi
Is it right?

 

[TeeJay]

Re: Toughest P1 Math questions I came across. A little help.

ds29 when you integrate the line y^2=9 u do it wid limits 2 and 0. in the diagram he has written 3 instead of 2 so u might have misread it. total volume will then com out 2 be 30pi!

 

[khizarr]

Re: Toughest P1 Math questions I came across. A little help.

teejay, it would be 33pi i think!
and SZ11 thanks a billion for this type of question ..

 

[TeeJay]

Re: Toughest P1 Math questions I came across. A little help.

How khizzar? I've double checked and the answer is still 30pi!

 

[SZ11]

Re: Toughest P1 Math questions I came across. A little help.

Yeah that's what I thought we're supposed to do. Thanks for helping out guys. Really appreciate it.
As for the trigonometric question, I apologise for the incoherency...
This link has the question: http://s941.photobucket.com/albums/ad25 ... 010048.jpg

Thanks again.

 

[SZ11]

Re: Toughest P1 Math questions I came across. A little help.

How khizzar? I've double checked and the answer is still 30pi!

Yeah. I checked too. 30pi it is.

 

[khizarr]

Re: Toughest P1 Math questions I came across. A little help.

yeah sorry, just a silly mistake, yeah its 30pi ..

 

[khizarr]

Re: Toughest P1 Math questions I came across. A little help.

for the identitty question, i dont think questions of this type would be included ... coz i did it somehow but i had to forst try with the right side ten left, when i reached at the same thing, with both the tries, i used the half solution of left side and proved it ..

 

[XWTBSILT]

Re: Toughest P1 Math questions I came across. A little help.

for the identitty question, i dont think questions of this type would be included ... coz i did it somehow but i had to forst try with the right side ten left, when i reached at the same thing, with both the tries, i used the half solution of left side and proved it ..

can u put it here

 

[XWTBSILT]

Re: Toughest P1 Math questions I came across. A little help.

I did the graph question
if u take x=3 and do u get 33pi
if u take x=2 and do u get 30pi
hehehehe

 

[lee91]

Re: Toughest P1 Math questions I came across. A little help.

the trigonometry question is really not easy!!!

there you go..

usually i'll prove it using both side, easier this way for complicated questions like this.

 

[lee91]

Re: Toughest P1 Math questions I came across. A little help.

how can i post picture here??? the uploader doesn't seem to work

 

[lee91]

Re: Toughest P1 Math questions I came across. A little help.

http://s979.photobucket.com/albums/ae27 ... G_9442.jpg

here it is

 

[XWTBSILT]

Re: Toughest P1 Math questions I came across. A little help.

Credit goes to the person who did this not me.
here it is

Let's deal with the numerator first

sin2x -cos2x

sin2x can be written as cos2x X tan 2x

that is because tanx=sinx/cosx

now the numerator looks like this

cos2x X tan 2x - cos2x

taking cos2x common you get

cos2x ( tan 2x - 1)

Now leave it at this. Move to the denominator

1+ 2sinxcosx
we know that sinx=cosxtanx

let's replace sinx we get

1+ 2 cosxtanx X cosx

1+ 2cos2xtanx

we also know
sin2x + cos2x =1

let's replace 1 with this

sin2x + cos2x + cos2xtanx

Now we still need to replace sin2x again by cos2x tan 2x

Doing so you get

cos2x tan 2x+ cos2x + 2cos2xtanx

Now take cos2x common

it reduces to
cos2x (1+ tan2x + 2tanx)

Now bring in the numerator again

cos2x ( tan 2x - 1)

divide the two
cos2x ( tan 2x -1)
----------------------------------------
cos2x (1+ tan2x + 2tanx)

cos2x gets cancelled

remaining bring it down again

( tan 2x - 1)-------> (tanx+1)(tanx-1) 1)
------------------------
(1+ tan2x + 2tanx)------>(tanx+1)2 2)

Divide 1 by 2

(tanx+1)(tanx-1)
-----------------
(tanx+1) (tanx+1)

(tanx+1)(tanx-1)
------------------------
(tanx+1) (tanx+1)

You have your answer
(tanx-1)
--------
(tanx+1)

 

[eddiegal]

Re: Toughest P1 Math questions I came across. A little help.

well i thnk tht identity can cum it was a bit tricky but i finally managed it in 3 attempts=P
dealing wid the numerator first sin^2x-cos^2x is in the form a^2-b^2 tht is (a+b)(a-b) nd therefore it becums (sinx+cosx)(sinx-cosx)
cumng to the denominator now 1+2sinxcosx..it can be expressed as sin^2x+cos^2x+2sinxcosx as sin^2x+cos^2x=1. this is a perfect square equation nd can be written as (sinx+cosx)^2 tht is (sinx+cosx)(sinx+cosx)
thus placing numerator over the denominator one of the (sinx+cosx) gets cancelled and we r left wid (sinx-cosx)/(sinx+cosx)
Divide the numeraotr nd denominator by cosx tht is [(sinx-cosx)/cosx]/[(sinx+cosx)/cosx]
and u will get (sinx/cosx -cosx/cosx)/(sinx/cosx+cosx/cosx) which gives u (tanx-1)/(tanx+1)
hope tht helped!=)