Question

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w13_qp_13.pdf

question 8,7,9 give me general idea how to do

Answer 1

if sin-1(x^2-1)=pi/3
then x^2-1=sin(pi/3)
x^2=sqrt(3)/2+1
now solve for x

Answer 2

what about the second part

Answer 3

use
sin-1(sina)=a
so
2(theta)+pi/3=sin-1(1/2)=pi/6
now solve for thete

Answer 4

ill get on it meanwhile can you tell how to do the question 8 third part

Answer 5

use this expnsion
(x + y)5 = x^5 +5x^4 y + 10x^3 y^2 +10x^2 y^3 +5xy^4 + y^5
http://www.mathsisfun.com/pascals-triangle.html

Answer 6

after i get teh coffeffecient there is a spefic way i am supposed to use it in third part
i just need clarification on how to do it

Answer 7

u just have to find all coeffs of x^8 and add them

Answer 8

in the first and second part ? for the third part ?

Answer 9

same procedure expand and add all coefficients of x^8

Answer 10

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w13_ms_13.pdf here is the mark scheme though

Answer 11

question 9 anyone ?

Answer 12

It has a stationary point where dy/dx = 0
so first find dy/dx