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shubhamsrg

if cos x + cos y + cos z = 3 sin60 and sin x + sin y + sin z = 3 cos60 then how do i prove that x=pi/6 + 2k(pi) y=pi/6 + 2l(pi) and z=pi/6 + 2m(pi) where k,l,m are integers..

  • one year ago
  • one year ago

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  1. vikrantg4
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    i dont have perfect solution.. still i logically say that cos x +cos y + cos z = cos 30 + cos 30 + cos 30 which means x=y=z=pi/6 since sin and cos repeat after intervals of 2pi x = y = z = pi/6 + 2n(pi)

    • one year ago
  2. vikrantg4
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    @shubhamsrg

    • one year ago
  3. shubhamsrg
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    i wont guess that'd be a good soln..thats just assumption..

    • one year ago
  4. vikrantg4
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    lol yeah .. :P

    • one year ago
  5. vikrantg4
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    @waterineyes

    • one year ago
  6. waterineyes
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    Thinking.. Ha ha ha.

    • one year ago
  7. shubhamsrg
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    @mukushla

    • one year ago
  8. waterineyes
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    offline.

    • one year ago
  9. shubhamsrg
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    i know,,just tagged him!! :P

    • one year ago
  10. shubhamsrg
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    maybe @UnkleRhaukus and @ganeshie8

    • one year ago
  11. shubhamsrg
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    i sq both and added them,,got cosx cosy + sinx siny + cosx cosz + sinx sinz + cosz cosy + sinz siny = 3 so its cos( ) + cos ( ) + cos ( ) = 3 where space inside ( ) denotes something now each cos ( ) = 1 but how do i reach my ans ?

    • one year ago
  12. shubhamsrg
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    @Callisto

    • one year ago
  13. waterineyes
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    \[\cos(x-y) + \cos(y-z) + \cos(z - x) = 3\]

    • one year ago
  14. vishweshshrimali5
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    well I recall something from \(2k\pi\) and that is demoirve's theorem from complex numbers which has a lot to do with such questions

    • one year ago
  15. waterineyes
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    I think you took it seriously @vishweshshrimali5

    • one year ago
  16. vishweshshrimali5
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    what ? @waterineyes

    • one year ago
  17. waterineyes
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    \(2k \pi\) is just coming from that fact that general solution of \(cos(x) = cos(y)\) is given by \[x = 2n \pi \pm y\]

    • one year ago
  18. vishweshshrimali5
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    yes......... trigonometric equations

    • one year ago
  19. waterineyes
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    Yes..

    • one year ago
  20. vishweshshrimali5
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    we can write sin x = cos(90-x)........ will that help ?

    • one year ago
  21. waterineyes
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    Let me remind the formula for : \((a + b + c)^2\)

    • one year ago
  22. vishweshshrimali5
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    Yes.............. \(\large a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\)

    • one year ago
  23. waterineyes
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    Okay..

    • one year ago
  24. vishweshshrimali5
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    Now ?

    • one year ago
  25. waterineyes
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    Check subhamarsh solution he is right in calculating this..

    • one year ago
  26. shubhamsrg
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    ????

    • one year ago
  27. waterineyes
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    Nothing..

    • one year ago
  28. mukushla
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    guys u already got it...

    • one year ago
  29. mukushla
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    \[\cos(x-y)+\cos(x-z)+\cos(y-z) \le1+1+1=3\]equality occurs when\[cos(x-y)=\cos(x-z)=\cos(y-z)=1\]

    • one year ago
  30. shubhamsrg
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    @mukushla how do we reach the ans from here?? also we dont know if it'll be x-y or y-x under ( ),,same argument for other ( )

    • one year ago
  31. waterineyes
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    That will not matter I think: \[\cos(-\theta) = \cos(\theta)\]

    • one year ago
  32. waterineyes
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    Oh wow....

    • one year ago
  33. waterineyes
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    I understand the feelings.. Ha ha ha...

    • one year ago
  34. shubhamsrg
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    sorry didnt catch ya ..come again..

    • one year ago
  35. shubhamsrg
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    x-y = 2npi ,, x-z = 2mpi,, after that..

    • one year ago
  36. mukushla
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    \[y=x-2n\pi\]\[z=x-2m\pi\] put them in the first one \[\cos x+\cos (x-2n\pi)+\cos(x-2m\pi)=3\cos 30\]\[\cos x+\cos x+\cos x=3\cos 30\]

    • one year ago
  37. shubhamsrg
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    aha..i see..thanks again..

    • one year ago
  38. mukushla
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    welcome

    • one year ago
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