Let us find values for cos x within the interval [0, pi]:

x=0 ==> Cos0 = 1

x= pi/2 = 3.14/2 = 1.57 ==> cospi/2 = 0

x= pi = 3.14 ==> cos pi= -1

Then we notice that the function is decreasin for the interval [0,pi]

==> cos3 < cos2 < cos 1

Then arranging increasingly would be:

cos3 cos2 cos1 (cos 1 is the largest and cos 3 is the smallest)

1 rad is in 1st quadrant . So cos (1rad)> 0 .

Cosx has a derivative -sinx. So cosx is decreasing in 1st and 2nd quadrant.

2 rad is in 2nd quadrant. cosx is decresing in second quadrant as /dx(cosx) = -sinx which is negative in 2nd quadrant as sign x is positive in 2nd quadrant.

Therefore cos(2rad) <cos(1rad)..

3 rad < pi , Cos x is decreasing in 2nd quadrant and miminimum at x = pi. So cos(3rad) < cos(2 rad).

Therefore , Cos(3rad) < (cos2rad) < (cos1rad).

If 1 ,2 and 3 are in degrees, cosx being a decresing in (1deg , 180 deg), Cos(3deg) < cos(2deg) < cos(1deg).

First, we'll establish the quadrants that contain the values,1 , 2, 3.

The first quadrant: (0 ;pi/2), where pi = 3.14, so pi/2 = 1.57

The second quadrant: (pi/2 ;pi) => (1.57;3.14)

It is obvious that 1 is in the interval (0,pi/2), meaning the first quadrant, where the function cosine is positive,then cos 1>0.

Because 2 and 3 are in the set (pi/2,pi), then cos2 and cos 3 are negative, because they belong to the second quadrant, where the values of cosine function are negative.

**So the biggest element is cos 1.**

**The numbers are arranged in this way: cos 3, cos 2, cos 1.**

For an angle in the first quadrant, cos values are positive; angles in the second and third quadrant have a negative cos value.

1 lies in the first quadrant therefore cos 1 is positive. 2 and 3 lie in the second quadrant where the value of cos of the angle decreases with an increase in the magnitude of the angle, 3 has a larger magnitude than 2.

Therefore cos 1> cos > cos3.