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 piarcos{cos(72π)}=72π

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 7/2pin72π

Note that if you rotate 2pi2π 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

7/2pi72π 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 2π radians =360o1 full cycle

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

So 7/2 pi 72π radians is (7/2 pi divide 2 pi)=7/4(72π÷2π)=74 cycles=1 3/4=134cycles.

Assuming you start at the standard point of 0^o,0o, you end up at the same point as3/434 cycles

In radian measure this becomes
3/4 times 360^o = 270^o = 3/4 times 2 pi =3/2 pi color(white)(.)34×360o=270o=34×2π=32π.radians. This is the same as -1/2 pi12π radians

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

Nov 21, 2015

arccos(cos((7pi)/2)) = pi/2arccos(cos(7π2))=π2

Explanation:

cos((7pi)/2) = cos(-pi/2) = cos(pi/2) = 0cos(7π2)=cos(π2)=cos(π2)=0

In general arccos(x)arccos(x) is an angle thetaθ
color(white)("XXX")XXXsuch that 0 <= theta < pi0θ<π
color(white)("XXX")XXXand cos(theta)= xcos(θ)=x

arccos((7pi)/2) = arccos(0) = pi/2arccos(7π2)=arccos(0)=π2