fp1 question

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  1. sabre2th1's Avatar
    • Exalted and Worshipped Member
    • Location: Southampton
    • Posts: 1,046
    fp1 question
    How do I write

    6(cos(3pi/4) + iSin(3pi/4)) in the form a + ib ?

    I can do it the other way, but this is harder..
  2. oli_G's Avatar
    • Vengeful, Imperial Overlord of The Student Room
    • Posts: 4,352
    Re: fp1 question
    (Original post by sabre2th1)
    How do I write

    6(cos(3pi/4) + iSin(3pi/4)) in the form a + ib ?

    I can do it the other way, but this is harder..
    By simple expansion:

     6(cos(\frac{3\pi}{4}) + isin(\frac{3\pi}{4})) = 6cos\frac{3\pi}{4} + 6sin\frac{3\pi}{4} i

    Can you get it from there?
  3. dknt's Avatar
    • Overlord in Training
    • Location: UK
    • Posts: 2,234
    Re: fp1 question
    (Original post by sabre2th1)
    How do I write

    6(cos(3pi/4) + iSin(3pi/4)) in the form a + ib ?

    I can do it the other way, but this is harder..
    It shouldn't be harder; if anything it's a lot easier. What is the value of cos(3pi/4) and sin(3pi/4)? Multiply by 6 and you there you have it!
  4. Fing4's Avatar
    • Exalted Member
    • Posts: 255
    Re: fp1 question
    (Original post by sabre2th1)
    How do I write

    6(cos(3pi/4) + iSin(3pi/4)) in the form a + ib ?

    I can do it the other way, but this is harder..
    Just work out what that comes to, it's not too hard

    (6cos(3pi/4)=a)

    (6sin(3pi/4)=b)
  5. David_Skiller's Avatar
    • Full Member
    • Posts: 126
    Re: fp1 question
    (Original post by GreenLantern1)
    De Moivre's Theorem
    :lol:
  6. f1mad's Avatar
    • TSR Demigod
    • Posts: 5,423
    Re: fp1 question
    (Original post by David_Skiller)
    :lol:
    Easy don .
  7. GreenLantern1's Avatar
    • Banned
    • Posts: 3,316
    • Warning points: 1000
    Re: fp1 question
    (Original post by David_Skiller)
    :lol:
  8. sabre2th1's Avatar
    • Exalted and Worshipped Member
    • Location: Southampton
    • Posts: 1,046
    Re: fp1 question
    (Original post by oli_G)
    By simple expansion:

     6(cos(\frac{3\pi}{4}) + isin(\frac{3\pi}{4})) = 6cos\frac{3\pi}{4} + 6sin\frac{3\pi}{4} i

    Can you get it from there?

    (Original post by dknt)
    It shouldn't be harder; if anything it's a lot easier. What is the value of cos(3pi/4) and sin(3pi/4)? Multiply by 6 and you there you have it!

    (Original post by Fing4)
    Just work out what that comes to, it's not too hard

    (6cos(3pi/4)=a)

    (6sin(3pi/4)=b)
    ah :facepalm: shoulda got that ! thankss
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