Turn on thread page Beta
    • Thread Starter
    Offline

    14
    ReputationRep:
    muh!

    someone help me out with question 3, ex 3c (edexcel)
    what i did was times out the brackets then rationalised the denominator but 'course im missing the = -1 bit


    if you dont do edexcel;

    Simply without usign a calculator:
    [cos(pi/7) - isin(pi/7)]^3 / [cos(pi/7) + isin(pi/7)]^4

    answer is

    cos(-pi) + isin(-pi) = -1
    Offline

    2
    ReputationRep:
    (Original post by kikzen)
    muh!

    someone help me out with question 3, ex 3c (edexcel)
    what i did was times out the brackets then rationalised the denominator but 'course im missing the = -1 bit


    if you dont do edexcel;

    Simply without usign a calculator:
    [cos(pi/7) - isin(pi/7)]^3 / [cos(pi/7) + isin(pi/7)]^4

    answer is

    cos(-pi) + isin(-pi) = -1
    E^(-3.i.pi/7)/E^(4.i.pi/7) = E^(-3ipi/7 + 4ipi/7) = E^-ipi = -1
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by It'sPhil...)
    E^(-3.i.pi/7)/E^(4.i.pi/7) = E^(-3ipi/7 + 4ipi/7) = E^-ipi = -1
    um ive never done that before; why did you do that?
    Offline

    2
    ReputationRep:
    (Original post by kikzen)
    um ive never done that before; why did you do that?
    e^ix = cosx + isinx . It makes problems like this a lot easier
    Offline

    2
    ReputationRep:
    (cos(pi/7)-isin(pi/7))^3/(cos(pi/7)+isin(pi/7))^4

    using (cosx+isinx)^n = cos(nx)+isin(nx),

    =(cos(3pi/7)-isin(3pi/7))/(cos(4pi/7)+isin(4pi/7))
    =(cos(3pi/7)-isin(3pi/7))(cos(4pi/7)-isin(4pi/7))/(cos^2(4pi/7)+isin^2(4pi/7))
    =cos(3pi/7)cos(4pi/7)-icos(3pi/7)sin(4pi/7)-isin(3pi/7)cos(4pi/7)-sin(3pi/7)sin(4pi/7)
    =cos(3pi/7+4pi/7)-isin(4pi/7+3pi/7)
    =cos(pi)-isin(pi)
    =-1 + 0
    =-1
    Offline

    2
    ReputationRep:
    (Original post by It'sPhil...)
    e^ix = cosx + isinx . It makes problems like this a lot easier
    The obvious alternative to the e^ix method is to multilpy top and bottom by (cos(pi/7) - isin(pi/7))^4
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by Fermat)
    (cos(pi/7)-isin(pi/7))^3/(cos(pi/7)+isin(pi/7))^4

    using (cosx+isinx)^n = cos(nx)+isin(nx),

    =(cos(3pi/7)-isin(3pi/7))/(cos(4pi/7)+isin(4pi/7))
    =(cos(3pi/7)-isin(3pi/7))(cos(4pi/7)-isin(4pi/7))/(cos^2(4pi/7)+isin^2(4pi/7))
    =cos(3pi/7)cos(4pi/7)-icos(3pi/7)sin(4pi/7)-isin(3pi/7)cos(4pi/7)-sin(3pi/7)sin(4pi/7)
    =cos(3pi/7+4pi/7)-isin(4pi/7+3pi/7)
    =cos(pi)-isin(pi)
    =-1 + 0
    =-1

    muh i didnt see the bit in bold so i was doing some dumb **** and ended up with a silly number on the bottom..

    cheers both of you !
    Offline

    10
    ReputationRep:
    (Original post by kikzen)
    muh i didnt see the bit in bold so i was doing some dumb **** and ended up with a silly number on the bottom..

    cheers both of you !
    Btw, you don't need to worry about the e^ix stuff, cos that's Euler's theorem (P5).
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by Nylex)
    Btw, you don't need to worry about the e^ix stuff, cos that's Euler's theorem (P5).
    ah no wonder
 
 
 
Poll
Do you think parents should charge rent?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.