[CIE] All Pure Mathematics (P1, P2 and P3) doubts here !

[elemis]

Thanks man,sorry for the trouble but i dont get Q6 part 1 in the same paper

Hold on.

[elemis]

If you have a complex number such as :

a + bi

Then, the modulus is :

For finding the argument of a complex no. :

Read your P3 textbook. There's an explanation in there.

[HUSH1994]

i already got that before,thanks man,but i have a prob with Q9 part b part 2

[elemis]

i already got that before,thanks man,but i have a prob with Q9 part b part 2

Okay...

[elemis]

Use the sum of an arithmetic progression formula and substitute :



You already know what the first term and the common difference is from (b)(i)

[HUSH1994]

i dont know why i post the question,i get the way right after i post em :S

[iluvme]

PR = PA + 0.5 AE + 0.5 AB      Read the information given in the question CAREFULLY

PQ = PO + OD + 0.5 DG

Oh right, Thanks again Ari.

Feeling very very stupid.

[Arthur Bon Zavi]

Oh right, Thanks again Ari.

Feeling very very stupid.

Any more doubts ?

[iluvme]

Any more doubts ?

Not right now.

Gotta start working again.

[Arissa_04]

Hey guys, need help in a trig question.not sure which year it is is from.Here's how the question goes:

Using the expansion of cos(3x-x) and cos(3x+x) prove that:

                    1/2(cos2x - cos4x) = sin3x sinx

Its a 3 mark question so I'm pretty sure its not that complicated but I cant seem to get the answers.
Thanks in advance

[ashish]

Hey guys, need help in a trig question.not sure which year it is is from.Here's how the question goes:

Using the expansion of cos(3x-x) and cos(3x+x) prove that:

                    1/2(cos2x - cos4x) = sin3x sinx

Its a 3 mark question so I'm pretty sure its not that complicated but I cant seem to get the answers.
Thanks in advance

cos(3x-x)= cos2x
             =cos3x * cosx -- sin3xsinx
             =cos3xcosx + sin3xsinx

cos(3x+x)=cos4x
             =cos3xcosx-sin3xsinx

cos2x - cos4x = cos3xcosx + sin3xsinx -(cos3xcosx - sin3xsinx)
                    = 2sin3xsinx

half of it is sin3xsinx
                     

[Arissa_04]

Thanks so much Ashish.
That was  well explained..I cant believe I spent so much time on that question unnecessarily complicating things by expanding cos3x and so on..
Thank you

[thecandydoll]

With respect to the origin O, the points A, B and C have position vectors given by
OA=i?k, OB=3i+2j?3k and OC=4i?3j+2k. The mid-point of AB is M. The point N lies on AC between A and C and is such that AN = 2NC.
(i) Find a vector equation of the line MN.   [4]
 (ii) It is given that MN intersects BC at the point P. Find the position vector of P.   [4]

I just dont get vectors sometimes,
any notes on this any notes on argand diagrams!
 thanks

[elemis]

With respect to the origin O, the points A, B and C have position vectors given by
OA=i?k, OB=3i+2j?3k and OC=4i?3j+2k. The mid-point of AB is M. The point N lies on AC between A and C and is such that AN = 2NC.
(i) Find a vector equation of the line MN.   [4]
 (ii) It is given that MN intersects BC at the point P. Find the position vector of P.   [4]

I just dont get vectors sometimes,
any notes on this any notes on argand diagrams!
 thanks

Long time since I've done any vector work.

AB = -i+k+3i+2j-3k
     = 2i -2k +2j

AM = 0.5 AB
     = i-k+j

OM = OA + AM
     = 2i +j -2k

AC = 3i -3j +3k

AN = (2/3)AC
    =  2i -2j +2k

ON = OA + AN
     = 3i -2j +k

Vector equation = position vector of N + (position vector of N - position vector of M)

Part (b)

Find the vector equation of BC :

r = 3i+2j-3k + (i -5j +5k)

Equate the equation from (a) with this new equation.

You will have three simultaneous equations.

Solve for lambda and mu. You should find lambda = 2 and mu =2

Re-insert back into one of the other equations and find the position vector of the cross over point.

[HUSH1994]

MJ 2004 Q10 part one,i was always wondering why do you have to change the sign of the least value,anyone knows that?