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rishabh.mission

matrix Question ....plz help

  • one year ago
  • one year ago

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  1. rishabh.mission
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    See the attachment

    • one year ago
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  2. rishabh.mission
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    please just solve this Question

    • one year ago
  3. phi
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    one way is use the trig identities \[ \tan \left(\frac{x}{2} \right)= \frac{1-\cos\ x}{\sin\ x }=\frac{\sin\ x }{1+\cos\ x}\]

    • one year ago
  4. rishabh.mission
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    plz buddy solve this i don't have any idea how to solve..

    • one year ago
  5. phi
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    you could start by writing down I-A can you do that ?

    • one year ago
  6. rishabh.mission
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    yes i did

    • one year ago
  7. rishabh.mission
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    I have problem with that equation in the circle i shown.

    • one year ago
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  8. phi
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    call 2 sin^2(a/2) = x then it looks like 1-x+x

    • one year ago
  9. phi
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    **look at the top left expression. Doesn't that simplify?

    • one year ago
  10. rishabh.mission
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    :( not understand

    • one year ago
  11. phi
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    or are you asking how they got to that point ?

    • one year ago
  12. rishabh.mission
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    yes

    • one year ago
  13. rishabh.mission
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    :)

    • one year ago
  14. rishabh.mission
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    that's Eq^n 1st

    • one year ago
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  15. phi
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    you just posted the left side. on the right side (the messy one) they did I-A times the other matrix I take it you understand how they got to the matrix labeled (2) ?

    • one year ago
  16. phi
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    the matrix above the circled stuff....

    • one year ago
  17. rishabh.mission
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    hmm

    • one year ago
  18. phi
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    We could go through that mess, but I would replace tan(a/2) with (1-cos)/sin in the top left entry: cos a + sin a tan(a/2) becomes cos a + sin a * (1-cos a)/sina= cosa + 1 -cosa = 1 which is what we want. Doing the way you have laid out is messier, but we can go through that if you want

    • one year ago
  19. rishabh.mission
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    can we write \[\cos \alpha/2 = (1-\tan ^{2} \alpha/2) / (1+\tan^{2} \alpha/2)\]

    • one year ago
  20. rishabh.mission
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    sorry ** its \[\cos \alpha \]

    • one year ago
  21. Hoa
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    can I say something?

    • one year ago
  22. rishabh.mission
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    yes please .. @hoa

    • one year ago
  23. Hoa
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    left side, i got I+A = \[\left[\begin{matrix}1 & -\tan \frac{ \alpha }{ 2} \\ \tan \frac{ \alpha }{ 2 } & 1\end{matrix}\right]\] = right side = (I-A) [ ] = \[\left[\begin{matrix}1 & \tan \frac{ \alpha }{ 2}\\ -\tan \frac{ \alpha }{ 2 } & 1\end{matrix}\right]\left[\begin{matrix}\cos & -\sin \\ \sin & \cos\end{matrix}\right]\]

    • one year ago
  24. Hoa
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    i just do the first term:

    • one year ago
  25. rishabh.mission
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    that's correct i already done this

    • one year ago
  26. Hoa
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    a11= cos + tanalpha/2* sin= 1

    • one year ago
  27. Hoa
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    you can expand the term like what phi says and you get 1 exactly what the left side matrix is

    • one year ago
  28. phi
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    for the top right of (2 - sin a + cos a tan(a/2) factor out -tan(a/2) (because that will be the answer) -tan(a/2) ( sin(a)/ tan(a/2) - cos(a)) replace 1/tan(a/2) inside the parens by multiplying by (1+cos(a))/sin a we get -tan(a/2) ( sin(a)*(1+cosa)/sina - cos(a))= -tan(a/2) ( 1 +cosa -cos a)= -tan(a/2) do the same thing for the other entries.

    • one year ago
  29. Hoa
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    because \[\cos + \sin *\tan \frac{ \alpha }{ 2 }= \cos + \sin\frac{ \sin }{ 1+\cos }= \frac{ \cos +\cos^2 +\sin^2 }{ 1+\cos}\] =1

    • one year ago
  30. Hoa
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    bingo. phi!! I am with you now. @rishabh.mission is it right?

    • one year ago
  31. rishabh.mission
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    Just last Query ... how both that matrix same ?

    • one year ago
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  32. Hoa
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    why do you make that question, why do you separate (I-A) and [cos sin....] matrix to calculate just the latter like that.?you may mess them up.

    • one year ago
  33. Hoa
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    they are a block and calculate together.

    • one year ago
  34. phi
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    I will do the top left entry 1- tan^2(a/2) is 1- sin^2(a/2)/cos^2(a/2) putting over a common denominator we get (cos^2(a/2) - sin^2(a/2)/cos^2(a/2) the numerator is cos(a) (one of version of the identity you need to know or look up) the denominator of the tp left entry 1+ tan^2(a/2) is 1+ sin^2(a/2)/cos^2(a/2) = (cos^2(a/2) + sin^2(a/2)/cos^2(a/2) = 1/cos^(a/2) the cos^2(a/2) cancels and we are left with cos a

    • one year ago
  35. phi
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    Can you follow that?

    • one year ago
  36. rishabh.mission
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    ok i done

    • one year ago
  37. rishabh.mission
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    @phi i can also use this cos(2a)=1-tan^2a/1+tan^2a

    • one year ago
  38. phi
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    that identity is true, but I have not tried to use it to simplify you problem. I tried to explain how I would do you problem.

    • one year ago
  39. rishabh.mission
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    ok you can solve this using this identity

    • one year ago
  40. phi
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    It does not look like a fruitful way to proceed, as you have sin and cos

    • one year ago
  41. rishabh.mission
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    No we can do using that Eq too

    • one year ago
  42. rishabh.mission
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    i'll show it to you in morning :)

    • one year ago
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