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..

Question

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..

anonymous · Answer

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)

anonymous · Answer

@shubhamsrg

shubhamsrg · Answer

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

anonymous · Answer

lol yeah .. :P

anonymous · Answer

@waterineyes

anonymous · Answer

Thinking..

Ha ha ha.

shubhamsrg · Answer

@mukushla

anonymous · Answer

offline.

shubhamsrg · Answer

i know,,just tagged him!! :P

shubhamsrg · Answer

maybe @UnkleRhaukus and @ganeshie8

shubhamsrg · Answer

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 ?

shubhamsrg · Answer

@Callisto

anonymous · Answer

$$\cos(x-y) + \cos(y-z) + \cos(z - x) = 3$$

vishweshshrimali5 · Answer

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

anonymous · Answer

I think you took it seriously @vishweshshrimali5

vishweshshrimali5 · Answer

what ? @waterineyes

anonymous · Answer

$2k \pi$ is just coming from that fact that general solution of $cos(x) = cos(y)$

is given by

$$x = 2n \pi \pm y$$

vishweshshrimali5 · Answer

yes......... trigonometric equations

anonymous · Answer

Yes..

vishweshshrimali5 · Answer

we can write sin x = cos(90-x)........ will that help ?

anonymous · Answer

Let me remind the formula for : $(a + b + c)^2$

vishweshshrimali5 · Answer

Yes.............. $\large a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

anonymous · Answer

Okay..

vishweshshrimali5 · Answer

Now ?

anonymous · Answer

Check subhamarsh solution he is right in calculating this..

shubhamsrg · Answer

????

anonymous · Answer

Nothing..

anonymous · Answer

guys u already got it...

anonymous · Answer

$$\cos(x-y)+\cos(x-z)+\cos(y-z) \le1+1+1=3$$equality occurs when$$cos(x-y)=\cos(x-z)=\cos(y-z)=1$$

shubhamsrg · Answer

@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 ( )

anonymous · Answer

That will not matter I think:

$$\cos(-\theta) = \cos(\theta)$$

anonymous · Answer

Oh wow....

anonymous · Answer

I understand the feelings..

Ha ha ha...

shubhamsrg · Answer

sorry didnt catch ya ..come again..

shubhamsrg · Answer

x-y = 2npi ,, x-z = 2mpi,,
after that..

anonymous · Answer

$$y=x-2n\pi$$$$z=x-2m\pi$$ put them in the first one
$$\cos x+\cos (x-2n\pi)+\cos(x-2m\pi)=3\cos 30$$$$\cos x+\cos x+\cos x=3\cos 30$$

shubhamsrg · Answer

aha..i see..thanks again..

anonymous · Answer

welcome