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shubhamsrg

  • 3 years ago

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..

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  1. vikrantg4
    • 3 years ago
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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)

  2. vikrantg4
    • 3 years ago
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    @shubhamsrg

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

  4. vikrantg4
    • 3 years ago
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    lol yeah .. :P

  5. vikrantg4
    • 3 years ago
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    @waterineyes

  6. waterineyes
    • 3 years ago
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    Thinking.. Ha ha ha.

  7. shubhamsrg
    • 3 years ago
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    @mukushla

  8. waterineyes
    • 3 years ago
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    offline.

  9. shubhamsrg
    • 3 years ago
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    i know,,just tagged him!! :P

  10. shubhamsrg
    • 3 years ago
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    maybe @UnkleRhaukus and @ganeshie8

  11. shubhamsrg
    • 3 years ago
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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 ?

  12. shubhamsrg
    • 3 years ago
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    @Callisto

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

  14. vishweshshrimali5
    • 3 years ago
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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

  15. waterineyes
    • 3 years ago
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    I think you took it seriously @vishweshshrimali5

  16. vishweshshrimali5
    • 3 years ago
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    what ? @waterineyes

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

  18. vishweshshrimali5
    • 3 years ago
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    yes......... trigonometric equations

  19. waterineyes
    • 3 years ago
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    Yes..

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

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

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

  23. waterineyes
    • 3 years ago
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    Okay..

  24. vishweshshrimali5
    • 3 years ago
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    Now ?

  25. waterineyes
    • 3 years ago
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    Check subhamarsh solution he is right in calculating this..

  26. shubhamsrg
    • 3 years ago
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    ????

  27. waterineyes
    • 3 years ago
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    Nothing..

  28. mukushla
    • 3 years ago
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    guys u already got it...

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

  30. shubhamsrg
    • 3 years ago
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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 ( )

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

  32. waterineyes
    • 3 years ago
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    Oh wow....

  33. waterineyes
    • 3 years ago
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    I understand the feelings.. Ha ha ha...

  34. shubhamsrg
    • 3 years ago
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    sorry didnt catch ya ..come again..

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

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

  37. shubhamsrg
    • 3 years ago
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    aha..i see..thanks again..

  38. mukushla
    • 3 years ago
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    welcome

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