The Edexcel C4 (21/06/12 - PM) Revision Thread

Can you show me what you did please...

Well, I think you've made a mistake in the question earlier on; I ended up with needing to simplify:

$5\ln \dfrac{5}{4}+2\ln \dfrac{4}{5}$

Using the rule I told you earlier:

$2\ln \dfrac{4}{5} = -2\ln \dfrac{5}{4}$

$\therefore 5\ln \dfrac{5}{4} - 2\ln \dfrac{5}{4} = 3\ln \dfrac{5}{4} = \ln \left( \left( \dfrac{5}{4}\right)^3\right)$

Do you know the answer?

someone please help me on q5d on the mock paper.. how do you prove the cartesian equation ? im so confused =[

heres the q :

x = cos t
y = sin 2t.
Expand: Sin2t = 2sin t Cos t
Now Substitute x = cos t.

therefore --> y = 2x sin t.

Use Sin^2 t + cos ^2 t = 1
Rearrange to get Sin t the subject.

Sin^2 t = 1 - cos^2 t
Substitute x = cos t
so Sin^2 t = 1- x^2
and Sin t= Sq.Root 1 - x^2

therefore goign back to the y = 2x Sin t

Substitute sin t as Sq.root 1 - x^2
Giving you the answer required.

Hope it helped.

Can someone explain what they will do if we talk about the exam before then?