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mathslover

  • 2 years ago

If \(\sin \theta + 2\sin \phi + 3\sin \Psi = 0 \) and \(\cos \theta + 2\cos \phi + 3\cos \Psi = 0\) ; then evaluate : a) \(\cos 3\theta + 8\cos 3 \phi + 27 \cos 3\Psi \) b) \(\sin (\phi + \psi) + 2\sin (\Psi + \theta) + 3 \sin (\theta + \psi)\)

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  1. mathslover
    • 2 years ago
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    @harsh314 @AravindG @blues @hartnn @Callisto @ghazi @jim_thompson5910

  2. mathsmind
    • 2 years ago
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    this question is solved, all you need is the sum of angles identity

  3. mathslover
    • 2 years ago
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    What I tried is : I did it for Part a) .. which is as follows Let \(\sin \theta + 2\sin \phi + 3\sin \Psi =0\) ------1) and \(\cos \theta + 2\cos \phi + 3\cos \Psi = 0\) be equation 2) Multiply equn 1) by iota (\(i\) ) ... and then add eqn 1) and 2) : we get : \(\cos \theta + i \sin \theta + 2\cos \phi + 2i\sin\phi + 3\cos \Psi + 3i \cos \Psi = 0\) Say \(\cos \theta + i \sin \theta =a \) , \(\cos \phi + i\sin \phi =b\) ,\(\cos \Psi + i \sin \Psi = c\) We get a + 2b + 3c = 0 then we get : \(a^3 + 8b^3 + 27c^3 = 18abc\) then simplifying it and equating the real parts I got : \(\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)]\)

  4. mathslover
    • 2 years ago
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    \[\large{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi) ]}\] ^ this is what I got for part a) .. I am having problem in part b)

  5. mathslover
    • 2 years ago
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    @dumbcow

  6. mathslover
    • 2 years ago
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    Got to go now but please answer this question if possible ... Thanks!

  7. mathslover
    • 2 years ago
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    Any one who can help me ??

  8. mathslover
    • 2 years ago
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    @dumbcow @UnkleRhaukus @hartnn @ghazi @ghass1978 @Callisto @jim_thompson5910 -- Any help or clue ?

  9. mathslover
    • 2 years ago
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    I am getting this when I tried Part 2) : \[\large{\sin 3\theta + 8\sin 3 \phi + 27 \sin 3 \psi = 18 \sin ( \theta + \phi + \psi )}\]

  10. mathslover
    • 2 years ago
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    OK leave it can any one prove that if : a + 2b + 3c = 0 then : \[\large{\frac{1}{2} + \frac{2}{b} + \frac{3}{c} = 0 }\]

  11. SamGracie
    • 2 years ago
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    the answer is SAM IS AMAZINGLY BEAUTIFUL no need to thank me ;D

  12. hartnn
    • 2 years ago
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    i think i've got something. To simplify writing, i'll take \(\theta =x , \phi =y, \psi =z, sin =s, cos =c.\) sx + 2sy = -3sz ----->(1) cx +2cy = -3cz----->(2) Square both (1) and (2) and then add. 1+ 4 [sxsy + cxcy ] + 4 = 9 4 [c (x-y)] = 4 c (x-y) = 1 x-y = 0 x=y. similarly if i take , sx + 3sz = -2sy ----->(1) cx + 3cz = -2cy----->(2) then u get x-z = \(\pi\) so, i have now, \(\theta = \phi = \pi +\psi \) substitute this in b, and continue.....

  13. hartnn
    • 2 years ago
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    not sure whether that would be useful...

  14. mathslover
    • 2 years ago
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    @hartnn can you help me in proving : 1/a + 2/b + 3/c = 0 if a + 2b + 3c = 0

  15. mathslover
    • 2 years ago
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    @shubhamsrg

  16. shubhamsrg
    • 2 years ago
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    a= cosx + isinx b= cosy + isiny c= cosz +isinz =>1/a = cosx - isinx 1/b= cosy -isiny 1/c = cosz -isinz Substitute and see if it helps ?

  17. mathslover
    • 2 years ago
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    1/a = cos x - i sin x / (cos^2 x - i sin ^2 x)

  18. experimentX
    • 2 years ago
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    i^2 = -1

  19. mathslover
    • 2 years ago
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    oh yep...

  20. mathslover
    • 2 years ago
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    i got it now ... thanks!

  21. mathslover
    • 2 years ago
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    I did like this : \[\large{a + 2b + 3c = 0}\] \[\large{ \bar {\bar{a}+2b+3\bar{c}} = 0}\] Which is : \[\large{e^{-i \theta} + 2e^{-i \phi} + 3e^{-i \Psi} = 0}\] Which I can write as : \[\large{\frac{1}{a} + \frac{2}{b} + \frac{3}{c} = 0}\] and then I get : \[\large{bc + 2ac + 3ab = 0}\] By simplifying this I get the required answer .... i.e. 0

  22. harsh314
    • 2 years ago
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    but i get the answer as\[\frac{ 9 }{ \sqrt{2} }\] for (a)

  23. mathslover
    • 2 years ago
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    how?

  24. harsh314
    • 2 years ago
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    at the end i got the required part as \[18(e ^{i (\alpha+ \beta+\gamma)}+e ^{-i(\alpha+\beta+\gamma)})\]

  25. mathslover
    • 2 years ago
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    yep than equate the real parts

  26. harsh314
    • 2 years ago
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    now solving it i \[18\cos(\alpha+\beta +\gamma)/2\]got \[\alpha +\beta +\gamma=\frac{ \pi }{ 4 }\] and hence the required answer

  27. harsh314
    • 2 years ago
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    \[e ^{i \times \theta}+e ^{-i \times \theta}=2 \cos(\theta)\]

  28. harsh314
    • 2 years ago
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    it is \[18 \frac{ \cos(\alpha +\beta+\gamma) }{ 2 }\]

  29. mathslover
    • 2 years ago
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    I think we have : \[{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi + i [\sin 3 \theta + \sin 3\phi +\\ \sin 3 \Psi ] = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)] }\]

  30. mathslover
    • 2 years ago
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    and then equating the real parts we get : \[\large{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi)]}\]

  31. mathslover
    • 2 years ago
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    Which mthod did you follow ? mine (mentioned on the top ) ? or any other ... ?

  32. harsh314
    • 2 years ago
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    yes the last portion is same what is the value of\[\alpha +\beta +\gamma\]

  33. harsh314
    • 2 years ago
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    i did it by using\[e ^{i \theta}+e ^{-i \theta}=2\cos(\theta),............e ^{i \theta}-e ^{-i \theta}=2i \sin \theta \]

  34. mathslover
    • 2 years ago
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    Well so how you got to this : \[\large{18 [e^{ i ( \theta + \phi + \psi)} + e^{-i (\theta + \phi +\psi)}]}\]?

  35. harsh314
    • 2 years ago
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    see the equations would be \[e ^{i \alpha}+ 2e^{i \beta}+3 e ^{i \gamma}=0\] and \[e ^{-i \alpha}+ 2e^{-i \beta}+3 e ^{-i \gamma}=0\]and from these we get \[e ^{2i \gamma}=e ^{i (\alpha+\beta)}\] cubing the first 2 equations and using the third one we get the mentioned thing

  36. mathslover
    • 2 years ago
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    you added the both equations ? : \[\large{e^{- i\alpha} + e^{i \alpha} + 2e^{-i \beta} + 2e^{i \beta} + 3e^{-i\gamma} + 3e^{i \gamma} = 0 }\] \[\large{2 ( \cos \alpha + 2 \cos \beta + 3 \cos \gamma )= 0}\]

  37. harsh314
    • 2 years ago
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    from the two given equations the two equations can be derived

  38. mathslover
    • 2 years ago
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    right so did I do right above .. ? and then : \[\large{ 2( \cos \alpha + 2\cos \beta + 3\cos \gamma )= 0}\] \[\large{ \cos \alpha + 2\cos \beta + 3 \cos \gamma = 0}\] Now since \(\cos \alpha + 2\ cos \beta + 3\ cos \gamma =0\) therefore : \[\large{(\cos \alpha)^3 + (2\cos \ beta )^3 + (3\cos \gamma)^3 = 3(\cos \alpha )(2\cos \beta) (3\cos \gamma)}\] \[\large{\cos 3 \alpha + 8 \cos 3 \beta +27 \cos 3 \gamma = 18 \cos \alpha \cos \beta \cos \gamma }\]

  39. harsh314
    • 2 years ago
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    no how did you get so

  40. mathslover
    • 2 years ago
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    I think I did wrong somewhere

  41. harsh314
    • 2 years ago
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    get the value of alpha +beta+gamma

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