IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => IB => Queries => Topic started by: candy on May 23, 2009, 09:47:26 am

Title: Pure 3 help...
Post by: candy on May 23, 2009, 09:47:26 am

please look at the attachment

Title: Re: Pure 3 help...
Post by: astarmathsandphysics on May 23, 2009, 10:01:25 am

sinx =o/h=a/f but since perimater of triangle is half perimeter of rectangle, 2r+rx=1/2*8a=4a so 4a=r(2+x) so r=4a/(2+x)

so sinx=a/r=a/(4a/(2+x))=1/4(2+x)

Title: Re: Pure 3 help...
Post by: candy on May 23, 2009, 10:34:47 am

Thanks astar!  ;)
r u giving math alevel? :)

Title: Re: Pure 3 help...
Post by: candy on May 23, 2009, 11:00:30 am

 ??? ??? part 2 please  ??? ???

Title: Re: Pure 3 help...
Post by: astarmathsandphysics on May 23, 2009, 01:25:14 pm

I am private Tutor. Wont be home for hours.

Title: Re: Pure 3 help...
Post by: louis on May 23, 2009, 04:18:48 pm

Candy,
The attachment I send contains solution for your question on complex number .

Title: Re: Pure 3 help...
Post by: candy on May 23, 2009, 07:42:09 pm

what program is this?  :-\
xml doesnt work on my pc  :(

Title: Re: Pure 3 help...
Post by: louis on May 24, 2009, 06:16:25 am

Candy,
Your question : Prove that the real part of 1 / ( Z+ 2 – i ) is a constant where
Z=2 cos a +i(1 -2 sin a )

Here is the solution :

1 / ( Z+ 2 – i )

= 1 / [ 2  cos a + i( 1 – 2 sin a) +2 –i ]

= 1 / [ 2 cos a +2 + i ( 1 – 2 sin a-1 )]

= 1  / ( 2 cos a+ 2 – 2 i sin a)

=(1/2)*   [ 1/ (cos a+ 1 – i sin a) ]

=(1/2)*   [ 1 + cos a + i sin a]  /   [ (1 + cos a) – i sin a] [( 1 + cos a) + i sin a ]

=(1/2)* [ 1 + cos a + i sin a] / [ (1 + cos a)² –( i sin a)² ]

=(1/2)* (1 + cos a+ i sin a ) /   [ 1 + 2 cos a+ cos² a+ sin² a ]

=(1/2)* (1 + cos a + i sin a ) / (  2 + 2 cos a )

=(1/4) *( 1 + cos a + i sin a ) / ( 1 +  cos  a)

=(1/4)  +   (1/4)*( i sin a)  / (  1 + cos a )

           

The real part is  1 / 4 which is a constant

Title: Re: Pure 3 help...
Post by: mac on May 24, 2009, 08:42:04 am

Another Pure maths 2/3 question...Please help...

For (ii) i find the angles by using a sin curve rather than +/-180 or 360.
I get confused by finding angles the other way! Do examiners accept the graphical method??
What are the methods that u guys use??
please Do the first part aswell... :)

 

Title: Re: Pure 3 help...
Post by: candy on May 24, 2009, 09:21:14 am

mac one way to remember the quadrant is : ALL(1st quadrant)...STUDENTS(2nd quadrant)...TAKE(3rd quadrant)...COFFEE(4th quadrant)

All= sin cos tan
Students: sin
Take: tan
Coffe: cos

see the attachment!  :) :)

Title: Re: Pure 3 help...
Post by: mac on May 24, 2009, 12:23:57 pm

thanks candy.
How can i do 2nd part....its not complet, its suppose to b 2x +67.4....
 

Title: Re: Pure 3 help...
Post by: louis on May 24, 2009, 12:50:43 pm

Candy,
I follow your working on the second part.I would like to forward the solution to you.
( My working below are in degree ,plz put the degree on yourself.)

13 sin ( 2x + 67.38 ) =11
 sin ( 2x + 67.38) = 11/13 = 0.8462
        2x + 67.38  =  57.8  degree,    122.2 degree
        2x              = -9.58,      54.82
                         = ( 360-9.580)  ,    54.82
                         =350.42,   54.82
          x              = 175.2,   27.41
                         =  175.2,  27.4             

Title: Re: Pure 3 help...
Post by: candy on May 24, 2009, 06:16:27 pm

ohh im sorry mac i misread the question...i didnt see the 2{theta} i read it as {theta} only
but i hope u understood the method!
Goodluck! :)

Title: Re: Pure 3 help...
Post by: mac on May 24, 2009, 06:47:12 pm

its ok i understand....i don't always get the angle right with this method....
Can i just use the sin graph to find the angles rather than +/- 180 or 360. I get confused with +/- , and when to use 180 or 360. What if i have to find more angles in the range. Do i just keep on adding 360??

 

Title: Re: Pure 3 help...
Post by: twilight on May 25, 2009, 07:32:18 pm

for frist quarant all the sin,cos and tan are positive
for second quadrant , only sine is positive
for third quadrant , only tan is positive
for fourth quadrant , only cosine is positive

and u havta add 360 to the reference angle .. which is 57.8 in this case .. not to the final answer