Question

Question 5ii pleasee!
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s07_qp_3.pdf

Answer 1

Previous ques, I got \[2 \cos \left( \theta -60 \right)\]

Answer 2

\[\int\limits _{0}^{\pi/2} \frac {d \theta}{4(\frac{1}{2}\cos \theta +\frac{\sqrt 3}{2} \sin \theta)^2}\]

Answer 3

isn't supposed to be related to i)

Answer 4

it is. look at the denominator carefully

Answer 5

\[\frac{1}{2}\cos \theta + \frac{\sqrt 3}{2} \sin \theta = \cos (60^o - \theta)\]

Answer 6

I don't get youre stating there. Where did you get your sin?

Answer 7

what*

Answer 8

in the given integral, divide and multiply the denominator by 4

Answer 9

\[4 = 2^2\]

Answer 10

Wait, can you slow down? You're too fast.

Answer 11

Can't we change it to sec?

Answer 12

yes, give me a minute I'll write down the first three steps

Answer 13

\[\int\limits^{\pi /2}_{0} \frac{d \theta}{(\cos \theta + \sqrt 3 \sin \theta)^2}=\int\limits^{\pi /2}_{0} \frac{d \theta}{\frac {4}{4}(\cos \theta + \sqrt 3 \sin \theta)^2} \]

Answer 14

why is it 4/4 ? why not 3/3 or 5/5

Answer 15

\[\int\limits^{\pi /2}_{0} \frac{d \theta}{4(\frac{1}{2}\cos \theta + \frac{\sqrt 3}{2} \sin \theta)^2}= \int\limits^{\pi /2}_{0} \frac{d \theta}{4(\cos 60^o \cos \theta + \sin 60^o \sin \theta)^2}\]

Answer 16

we chose 4 because we needed 1/2 and sqrt(3)/2 inside the bracket

Answer 17

why?

Answer 18

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s07_ms_3.pdf Take a look at the answer scheme. It's not that complicated D:

Answer 19

to use part i -->cos (60- theta)
and since sqrt(3) was already there i just needed the denominator to be 2

Answer 20

\[ 4(\frac{1}{2}\cos \theta + \frac{\sqrt 3}{2} \sin \theta)^2= 4(\cos \theta \cos 60^o + \sin 60^o \sin \theta)^2= 4 [\cos (\theta - 60^o)]^2\]

Answer 21

\[\int\limits^{\pi /2}_{0} \frac{d \theta} {4 \cos^2 (\theta -60^o)} = \int\limits^{\pi /2}_{0}\frac{1}{4} \times \sec^2 (\theta - 60^o) d \theta\]

Answer 22

now we have arrived at the first step of the marking scheme

Answer 23

converted the given integral in to sec^2 theta-60

Answer 24

Omg. im so lost. :(

Answer 25

don't worry. lets try another approach

Answer 26

Okay.

Answer 27

I think you should see my workings first.

Answer 28

how did you figure out what to do for 5i?

Answer 29

yes, please put it up

Answer 30

Erm for 5i it's rather easy, i just follow the formula. then i got 2cos(theta-60)

Answer 31

as for ii, i wass trying to expand the denominator, i got 1/ (4cos^2 theta - 4cos theta +1)

Answer 32

you mean you worked your way back for 5i? how did you get 2?

Answer 33

we'll get to 5ii in time, lets talk about part 5i first

Answer 34

R = sqrt(a^2+b^2)? It's just the formula.

Answer 35

correct. this formula comes from pythagoras theorem.

Answer 36

i have the problem solving ii only. D:

Answer 37

think about why the formula works and you'll know why i used 4/4 and not 3/3/ or 5/5.