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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..

Mathematics
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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)
i wont guess that'd be a good soln..thats just assumption..

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Other answers:

lol yeah .. :P
Thinking.. Ha ha ha.
offline.
i know,,just tagged him!! :P
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 ?
\[\cos(x-y) + \cos(y-z) + \cos(z - x) = 3\]
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
I think you took it seriously @vishweshshrimali5
\(2k \pi\) is just coming from that fact that general solution of \(cos(x) = cos(y)\) is given by \[x = 2n \pi \pm y\]
yes......... trigonometric equations
Yes..
we can write sin x = cos(90-x)........ will that help ?
Let me remind the formula for : \((a + b + c)^2\)
Yes.............. \(\large a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\)
Okay..
Now ?
Check subhamarsh solution he is right in calculating this..
????
Nothing..
guys u already got it...
\[\cos(x-y)+\cos(x-z)+\cos(y-z) \le1+1+1=3\]equality occurs when\[cos(x-y)=\cos(x-z)=\cos(y-z)=1\]
@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 ( )
That will not matter I think: \[\cos(-\theta) = \cos(\theta)\]
Oh wow....
I understand the feelings.. Ha ha ha...
sorry didnt catch ya ..come again..
x-y = 2npi ,, x-z = 2mpi,, after that..
\[y=x-2n\pi\]\[z=x-2m\pi\] put them in the first one \[\cos x+\cos (x-2n\pi)+\cos(x-2m\pi)=3\cos 30\]\[\cos x+\cos x+\cos x=3\cos 30\]
aha..i see..thanks again..
welcome

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