How do you evaluate #arccos(cos(5pi/4))#?

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Answer 1 · 2015-07-29T14:50:20

Evaluate #arccos (cos ((5pi)/4))#
Ans:# +- (3pi)/4#

#cos ((5pi)/4) = cos (pi/4 + pi) = - cos pi/4 = -(sqrt2)/2#

#arc x = arccos ((-sqrt2)/2)# ---> #x = +- (3pi)/4#

Answer 2 · 2015-07-29T15:56:23

#(3pi)/4#

#arccosx# can be thought of as an angle that measures between #0# and #pi# radians whose cosine is x.

(It can also be thought of as simply a number between #0# and #pi# whose cosine is #x#.)

The restriction to angles between #0# and #pi# makes #arccos# a function.

#arccos(cos((5pi)/4))# is an angle between #0# and #pi# whose cosine is the same as the cosine of #(5pi)/4#.

The angle we want is #(3pi)/4#

We know that #cos((5pi)/4) = -sqrt2/2#
and the Quadrant II angle with cosine equal to #-sqrt2/2#
is #(3pi)/4#