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if sinx + siny + sinz =3 find cosx +cosy + cosz?
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Maximum value of sinθ is 1.
So if sinx + siny + sinz = 3, then sinx = siny = sinz = 1
This does not mean that each of x, y, z = π/2
[There are infinitely many values of x that give sinx = 1]
But it does mean that cosx = cosy = cosz = 0
sin²x + cos²x = 1
1² + cos²x = 1
cos²x = 0
cosx = 0
Similarly, we get cosy = 0, cosz = 0
cosx + cosy + cosz = 0 + 0 + 0 = 0
For all x, -1 <= sinx <= 1, so if sinx + siny + sinx = 3 then sinx = siny = sinz = 1. Now, since sin²x + cos²x = 1, cosx = cosy = cosz = 0 and
cosx + cosy + cosz = 0
Hi,
the maximum value of sina is 1. That means that cosa =0.
If sinx+siny+sinz=3 cosx+cosy+cosz=0.
-1≤sinx≤1 so sinx+siny+sinz=3 requires
sinx=1, siny=1, sinz=1 so
x=(2m+1)pi/2, y=(2n+1)pi/2, z=(2r+1)pi/2 where m,n,r are integers.
So cosx+cosy+cosz=cos(2m+1)pi/2+cos(2n+1)pi/2+cos(2r+1)pi/2
=0+0+0
=0.
A bit longerr than Sir's answer.
0
sinx+siny+sinz = 3 iff sinx=siny=sinz=1 which , in turn, implies cosx=cosy=cosz = 0.
Since maximum value of sinA = 1
sin90 = 1
Hence cos90 = 0
cosx + cosy + cosz = 0
x=y=z
3sinx=3 -> sinx=1
sin²x=1
sin²x=1-cos²x ->
1-cos²x=1
cos²x=0
cosx=0
-> cosx+cosy+cosz=0
The maximum of sin x , sin y and sinz= 1 each. Therefore x= y = z= pi/2. Therefore, cos x= cos y= cos z=0. The sum is 0.
cos x + cos y + cos z = 0 when x = pi / 2
Since, sin x + sin y + sin z = 3 when x = pi / 2
Note: sin ( pi/ 2) = 1 and cos (pi/2) = 0