fbpx

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

    ##(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##