How do you evaluate #cos^(-1) [cos((-5pi)/3)]#?

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Answer 1 · 2017-07-21T11:15:41

#(-5pi)/3#

#cos^-1[cos( (-5pi)/3)]#

Let #cos^-1[cos( (-5pi)/3)] = theta# , then

#cos theta = cos( (-5pi)/3)#

#:. theta = (-5pi)/3#

#:.cos^-1[cos( (-5pi)/3)] = (-5pi)/3# [Ans]

Answer 2 · 2017-07-22T05:23:47

#cos^-1(cos((-5pi)/3)) = pi/3#

Here's the trick to this problem: #cos^-1(theta)# is only defined on the interval #0 le theta le pi#.

So, what this problem is asking us to do is find the angle between #0# and #pi# whose cosine is the same as the cosine of #(-5pi)/3#.

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Adding #2pi# to an angle doesn't change any of its trig functions, so we can say that:

#cos((-5pi)/3) = cos((-5pi)/3 + 2pi)#

#cos((-5pi)/3) = cos(pi/3)#

And since #pi/3# is between #0# and #pi#, we can say that:

#cos^-1(cos((-5pi)/3)) = pi/3#

Final Answer