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 $(3pi)/4$

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