How do you calculate Arccos(cos 7pi/2)?

2 Answers
Nov 21, 2015

arcos(cos) cancel each other out so #"arcos"{cos(7/2 pi)}=7/2 pi#

Explanation:

The problem with trig is that unless the range is defined there are multiple answers as the cycles just keep on going. So technically it should be written as #n 7/2pi#

Note that if you rotate #2pi# radians you have completed 1 cycle and you are back where you started. I suppose that if you are talking about work done (physics) the number of rotations is important. However, if you are talking purely numbers then it is not so critical

#7/2pi# radians is actually defining the number of rotations. However, just being interested in the angular measure of rotational state you would have to take into account that #2pi" radians "= 360^o -= 1 # full cycle

#color(brown)("Assumption: rotation is anticlockwise:")#

So #7/2 pi # radians is #(7/2 pi divide 2 pi)=7/4# cycles#=1 3/4#cycles.

Assuming you start at the standard point of #0^o,# you end up at the same point as#3/4# cycles

In radian measure this becomes
#3/4 times 360^o = 270^o = 3/4 times 2 pi =3/2 pi color(white)(.)#radians. This is the same as# -1/2 pi# radians

#color(blue)("In this solution I have assigned negative to be")#
#color(blue)("clockwise rotation and obviously positive to be anticlockwise")#

Nov 21, 2015

#arccos(cos((7pi)/2)) = pi/2#

Explanation:

#cos((7pi)/2) = cos(-pi/2) = cos(pi/2) = 0#

In general #arccos(x)# is an angle #theta#
#color(white)("XXX")#such that #0 <= theta < pi#
#color(white)("XXX")#and #cos(theta)= x#

#arccos((7pi)/2) = arccos(0) = pi/2#