Qualification > Math

Pure math(p3)

<< < (3/9) > >>

ITS MEE...:
oops it looks so awkward but its nove 2008 p3 CIE

astarmathsandphysics:

--- Quote from: preity on May 28, 2009, 08:56:23 am ---Can someone help me with this questions

M/J 02 Q1
M/J 03 Q1
0/N 04 Q4

--- End quote ---

M/J 02 Q1

cotx-tanx=cosx/sinx-sinx/cosx
=((cosx)^2-(sinx)^2)/sinxcosx=cos2x/(1/2sin2x)=2cot2x

M/J 03 Q1

sin(x ? 60 ) ? cos(30 ? x) = 1
sinxcos60-cosxsin60-cos30cosx-sin30sinx=1
sinx*1/2-cosx-cosx-1/2sinx=1
cosx=1 so cosx=1/

0/N 04 Q4
tan(45 + x) = 2 tan(45 ? x)

(tan45+tanx)/(1-tan45tanx)=2(tan45-tanx)/(1tan45tanx)


(1+tanx)/(1-tanx)=2(1-tanx)/(1+tanx)
cross multiply
(1+tanx)^2=2(1-tanx)(1-tanx)
1+2tanx+tan^2x=2-4tanx+2tan^2x
tan^2x-6tanx+1=0

ITS MEE...:
OK can someone do
nov08 question 7)ii
                     10)iv
                     8)iii
thank u

ITS MEE...:
please someone answer ???

twilight:

--- Quote from: preity on May 28, 2009, 08:56:23 am ---Can someone help me with this questions

M/J 02 Q1
M/J 03 Q1
0/N 04 Q4

--- End quote ---
for 0/N 04 Q4
(i) tan(45+x) = 2tan(45-x)                use tan(a+b)= (tan a + tan b)/(1- tan a tan b )     in the formula sheet
   (tan 45  +  tan x)/(1 -  tan 45  tan x) = 2(tan 45  - tan x)/(1 +  tan 45 tan x )
(1+tan x )/(1-tan x) = (2-2tan x)/(1+tan x)
cross multiply to get  tan2x - 6tanx + 1 = 1

(ii) substitute tanx by y
then solve by y=   -b+(b2-4ac)
                       ________________
                                  2a
then y =   -b-(b2-4ac)
            _________________
                       2a

then substitute back with tanx and do the usual method to get the values for x for + and -

Navigation

[0] Message Index

[#] Next page

[*] Previous page