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.