Exact value of cos(π÷12)

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  • เผยแพร่เมื่อ 23 ธ.ค. 2024

ความคิดเห็น • 42

  • @PauxloE
    @PauxloE 3 ปีที่แล้ว +11

    3:26 "Two things happen when you multiply by a complex number: [...] you scale, that comes from the real component, and you rotate, which comes from the imaginary component."
    No, just no. Multiplying by 2i (which has no real component) is still scaling by 2 (besides the rotation by pi/2).
    It's the modulus (or absolute value) which is scaling, and the argument which is rotating.

  • @MarkMifsud
    @MarkMifsud 3 ปีที่แล้ว

    I never get tired of this kind of teaching. I plan to work a lot of problems to master these skills.

  • @damianbla4469
    @damianbla4469 3 ปีที่แล้ว +1

    12:40 You could also use the trigonometric identity
    cos^2(x) + sin^2(x) = 1
    as
    cos^2(x) - i^2 * sin^2(x) =
    = cos^2(x) - (-1) * sin^2(x) =
    = cos^2(x) - (-sin^2(x)) =
    = cos^2(x) + sin^2(x) =
    = 1
    No matter which angle is "x"
    (in our case it was PI/4),
    the result is equal to 1.
    P.S.
    The identity
    cos^2(x) + sin^2(x) = 1
    in the imaginary form:
    cos^2(x) - i^2 * sin^2(x) = 1
    or
    [cos(x) + i * sin(x)] * [cos(x) - i * sin(x)] = 1

    • @reynep
      @reynep ปีที่แล้ว

      ahaaaa That's pretty nice.

  • @MrBennedy
    @MrBennedy 3 ปีที่แล้ว +7

    15:42 tiny correction: *plus* i sin pi/12

    • @matemaatika-math
      @matemaatika-math 3 ปีที่แล้ว

      Why?

    • @MrBennedy
      @MrBennedy 3 ปีที่แล้ว +1

      @@matemaatika-math It's a small slip up that can tend to happen at the end of a long piece of working. Presumably, he had the word "minus" on his mind because he just was just referencing the pi/3 *minus* pi/4. Commentating on what he had just written possibly made him write minus instead of plus. I hope this answers your wondering of why. If you're asking why in terms of mathematics - because the final line purports that cos(pi/12) - i sin (pi/12)= cos(pi/12) + i sin (pi/12), which is not true. It doesn't have a effect on the answer of the final question because the the real part of the expression is still correct. I'm certain that had been live in a classroom, one of the students would have pointed the small copying error out.

    • @matemaatika-math
      @matemaatika-math 3 ปีที่แล้ว

      @@MrBennedy You're right. Thank you!

  • @marciocastro7101
    @marciocastro7101 3 ปีที่แล้ว +2

    About denominator with cosine squared of pi over 4 plus sine squared of pi over 4, is possible to use the Fundamental Pythagorean Identity...

  • @pjmmccann
    @pjmmccann 3 ปีที่แล้ว +7

    An errant "-" instead of a "+" creeps in at 15:38 , but it doesn't affect the answer.

    • @matemaatika-math
      @matemaatika-math 3 ปีที่แล้ว

      Why do you count it as an errant? He explained what formula he used.

    • @rohitchaoji
      @rohitchaoji 3 ปีที่แล้ว +2

      @@matemaatika-math It's errant because the sign of the sine function should not change. sin (pi/3 - pi/4) is still positive sin(pi/12), but like he said, since we're dealing with just the real part, it didn't matter, and probably why Eddie did not notice it either. He surely would have if that part were important.

  • @沈博智-x5y
    @沈博智-x5y 3 ปีที่แล้ว

    12:29
    This is just cos^2(x) + sin^2(x) = 1 , the pythagorean identity
    no need for exact values; although it is reassuring we get 1 anyway.
    Let conj(z) be the conjugate of z
    z*conj(z) = |z|^2
    if z = cos(x) + isin(x)
    Then z*conj(z) = |z|^2 = (sqrt(cos^2(x) + sin^2(x)))^2 = (sqrt(1))^2 = 1

  • @jamesseth8149
    @jamesseth8149 3 ปีที่แล้ว

    eddie woo one of the best asian youtube teacher 😄

  • @damianbla4469
    @damianbla4469 3 ปีที่แล้ว

    18:15 "(1 + sqrt(3)) / (2*sqrt(2))" - now you could multiply both nominator and denominator by sqrt(2) to get
    cos(PI/12) = (sqrt(2) + sqrt(6)) / 4

  • @grahamdoesmaths
    @grahamdoesmaths 3 ปีที่แล้ว

    Lovely work. The 1 in the denominator was a delight 🤗

  • @eliabtesfaye7436
    @eliabtesfaye7436 3 ปีที่แล้ว

    Question: to realise the denominator, instead of multiplying the top and bottom of the fraction by the denominators conjugate, couldn’t you just multiply the top and bottom by ‘i’ ?

  • @nithinkondababathini6001
    @nithinkondababathini6001 ปีที่แล้ว +1

    You can use Cis notation

  • @andreinecula9492
    @andreinecula9492 3 ปีที่แล้ว +1

    Did this topic recently, a bit of revision won't do any harm :)

  • @Seppevh
    @Seppevh 3 ปีที่แล้ว +3

    Question B would be made so much easier with exponential form XD

    • @matemaatika-math
      @matemaatika-math 3 ปีที่แล้ว

      What form is that?

    • @Seppevh
      @Seppevh 3 ปีที่แล้ว

      @@matemaatika-math The form z=re^(iθ)

    • @matemaatika-math
      @matemaatika-math 3 ปีที่แล้ว

      @@Seppevh I'm not aware of that and pretty sure that this form isn't in the curriculum.

    • @Seppevh
      @Seppevh 3 ปีที่แล้ว

      ​@@matemaatika-math Its related to the Mod-arg form used in the video since e^iθ = cosθ + isinθ.
      Using it is pretty much doing the same thing as he mentions at 15:45 where you can just divide the modulo and substract the angles.

    • @yuvalnachum4190
      @yuvalnachum4190 3 ปีที่แล้ว +1

      if you know r1cisA/r2cisB=r1/r2cis(A-B) it should take 5 sec

  • @GooogleGoglee
    @GooogleGoglee 3 ปีที่แล้ว

    Astonishing...

  • @johnsavard7583
    @johnsavard7583 3 ปีที่แล้ว

    That would be the cosine of 30 degrees. Its sine is 1/2, so the pythagorean theorem gives you half of the square root of 3 as the answer. I take it you're doing it the hard way.

  • @evansokolson9221
    @evansokolson9221 3 ปีที่แล้ว +1

    The exact value of cos(п/12) is cos(п/12)

  • @aayushjha5004
    @aayushjha5004 3 ปีที่แล้ว +1

    Sir how can I become a great mathsmatician like you.

  • @vineet5701
    @vineet5701 3 ปีที่แล้ว

    We learn this in class 10😅😅

    • @particleonazock2246
      @particleonazock2246 3 ปีที่แล้ว

      Where Eddie comes from, we learn this in Year 11/12.

  • @MovieStuff-ug9cq
    @MovieStuff-ug9cq 3 ปีที่แล้ว

    5th

  • @Piyush.91
    @Piyush.91 3 ปีที่แล้ว

    4th

  • @MovieStuff-ug9cq
    @MovieStuff-ug9cq 3 ปีที่แล้ว

    ha
    ha

  • @MovieStuff-ug9cq
    @MovieStuff-ug9cq 3 ปีที่แล้ว

    h

  • @MovieStuff-ug9cq
    @MovieStuff-ug9cq 3 ปีที่แล้ว

    h
    h
    h
    h

  • @MovieStuff-ug9cq
    @MovieStuff-ug9cq 3 ปีที่แล้ว

    i

  • @MovieStuff-ug9cq
    @MovieStuff-ug9cq 3 ปีที่แล้ว

    h

  • @MovieStuff-ug9cq
    @MovieStuff-ug9cq 3 ปีที่แล้ว

    h