Hey there! Sign in to join this conversationNew here? Join for free
Turn on thread page Beta
    Offline

    20
    ReputationRep:
    About to do IAL S15, wish me luck!
    Offline

    3
    ReputationRep:
    Hi, Please could someone do a worked solution to question 12a please june 2004 paper https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf as I don't really understand the method on the mark scheme, thank you
    Online

    22
    ReputationRep:
    (Original post by economicss)
    Hi, Please could someone do a worked solution to question 12a please june 2004 paper https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf as I don't really understand the method on the mark scheme, thank you
    Is there a specific but that you don't understand about the markscheme? Have you looked at examsolutions.net videos on complex transformations?
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    Is there a specific but that you don't understand about the markscheme? Have you looked at examsolutions.net videos on complex transformations?
    Yeah, I understand how you get to the fraction representing w but I don't really understand how you'd know to go from here to finding the modulus of the numerator and denominator as I've not seen this method used before? Yeah I have, but I'm still struggling with bits of them quite a lot :/ Thanks
    Offline

    12
    ReputationRep:
    Can someone please help me with this step? I don't understand where the (pie/3 - theta ) goes

    btw it is from Jan05 p4 Q7
    Thanksbtw it
    Attached Images
     
    Offline

    3
    ReputationRep:
    Name:  image.jpg
Views: 92
Size:  449.3 KB
    (Original post by Cpj16)
    Can someone please help me with this step? I don't understand where the (pie/3 - theta ) goes

    btw it is from Jan05 p4 Q7
    Thanksbtw it
    Not sure if this helps, but this is the working I did for the question
    Offline

    12
    ReputationRep:
    (Original post by economicss)
    Name:  image.jpg
Views: 92
Size:  449.3 KB
    Not sure if this helps, but this is the working I did for the question
    thanks , I do understand this (this was part a of the question )

    if possible could you explain the step I attached earlier

    edit:the image is part of the mark scheme and is not a question I just don't understand how they got there
    Offline

    3
    ReputationRep:
    (Original post by Cpj16)
    thanks , I do understand this (this was part a of the question )

    if possible could you explain the step I attached earlier

    edit:the image is part of the mark scheme and is not a question I just don't understand how they got there
    Sure, is it part c you'd like a worked solution to?
    Offline

    12
    ReputationRep:
    (Original post by economicss)
    Sure, is it part c you'd like a worked solution to?
    yes please that would be really helpful
    Online

    22
    ReputationRep:
    (Original post by Cpj16)
    Can someone please help me with this step? I don't understand where the (pie/3 - theta ) goes

    btw it is from Jan05 p4 Q7
    Thanksbtw it
    \displaystyle \cos \left(\frac{\pi}{3}  - \theta\right) = \frac{1}{2}\cos \theta + \frac{\sqrt{3}}{2}\sin \theta

    So \displaystyle 6\cos \theta \left(\frac{1}{2}(\cos \theta + \sqrt{3}\sin \theta\right) = 3\cos^2 \theta + 3\sqrt{3}\sin \theta \cos \theta

    Therefore: \cos^2 \theta + \sqrt{3}\sin \theta \cos \theta = 1 \iff \cdots
    Offline

    12
    ReputationRep:
    (Original post by Zacken)
    \displaystyle \cos \left(\frac{\pi}{3}  - \theta\right) = \frac{1}{2}\cos \theta + \frac{\sqrt{3}}{2}\sin \theta

    So \displaystyle 6\cos \theta \left(\frac{1}{2}(\cos \theta + \sqrt{3}\sin \theta\right) = 3\cos^2 \theta + 3\sqrt{3}\sin \theta \cos \theta

    Therefore: \cos^2 \theta + \sqrt{3}\sin \theta \cos \theta = 1 \iff \cdots
    thank you!! I would rate you up but tsr is telling to not rate the same member again
    Offline

    3
    ReputationRep:
    (Original post by Cpj16)
    thanks , I do understand this (this was part a of the question )

    if possible could you explain the step I attached earlier

    edit:the image is part of the mark scheme and is not a question I just don't understand how they got there
    Name:  image.jpg
Views: 86
Size:  453.7 KB
    Offline

    12
    ReputationRep:
    (Original post by economicss)
    Name:  image.jpg
Views: 86
Size:  453.7 KB
    Thank you!!!
    Online

    22
    ReputationRep:
    (Original post by Cpj16)
    thank you!! I would rate you up but tsr is telling to not rate the same member again
    Nobody seems to be able to rep me, conspiracy. :erm:

    jk, no problem!
    Offline

    1
    ReputationRep:
    (Original post by edothero)
    I'm half-way through TeeEm's booklet, some really really tough questions at the end..
    where can I find this booklet!! need some extra revision material.
    thanks
    and good luck with your studies
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    Is there a specific but that you don't understand about the markscheme? Have you looked at examsolutions.net videos on complex transformations?
    It's probably obvious but on the MS how do you know that the square root of lambda+1 squared plus lambda squared is 1? Thanks
    Online

    22
    ReputationRep:
    (Original post by economicss)
    It's probably obvious but on the MS how do you know that the square root of lambda+1 squared plus lambda squared is 1? Thanks
    It's not saying the square root of the thingies squared is 1. It's saying that w = \frac{\lambda + 1 + \lambda i}{\lambda + (\lambda +1)i}

    So that \displaystyle |w| = \left| \frac{\lambda + 1 + \lambda i}{\lambda + (\lambda +1)i}\right| =  \frac{|\lambda + 1 + \lambda i|}{|\lambda + (\lambda +1)i|} = \frac{\sqrt{(\lambda+1)^2 + \lambda^2}}{\sqrt{\lambda^2 + (\lambda + 1)^2}} = 1

    i.e: the modulus of the numerator is the same as the modulus of the denominator. Hence the modulus of the entire thing is just 1.
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    It's not saying the square root of the thingies squared is 1. It's saying that w = \frac{\lambda + 1 + \lambda i}{\lambda + (\lambda +1)i}

    So that \displaystyle |w| = \left| \frac{\lambda + 1 + \lambda i}{\lambda + (\lambda +1)i}\right| =  \frac{|\lambda + 1 + \lambda i|}{|\lambda + (\lambda +1)i|} = \frac{\sqrt{(\lambda+1)^2 + \lambda^2}}{\sqrt{\lambda^2 + (\lambda + 1)^2}} = 1

    i.e: the modulus of the numerator is the same as the modulus of the denominator. Hence the modulus of the entire thing is just 1.
    Ah yes ofc, I'm being stupid! Thanks for your help
    Online

    22
    ReputationRep:
    (Original post by economicss)
    Ah yes ofc, I'm being stupid! Thanks for your help
    No worries.
    Offline

    1
    ReputationRep:
    For de moivre's theorem, do we need to know the proof for both positive and negative integers? I mean positive i can manage it's just proof by induction by negative is just like extraaa...
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 24, 2017
Poll
“Yanny” or “Laurel”

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.