How do you evaluate $arccos (cos (5 \frac{π}{4}))$ $arccos(cos(5pi/4))$ ?

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How do you evaluate $arccos (cos (5 \frac{π}{4}))$ $arccos(cos(5pi/4))$ ?

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Answer 1 · 2015-07-29T14:50:20

Evaluate $arccos (cos (\frac{5 π}{4}))$ $arccos (cos ((5pi)/4))$
Ans: $\pm \frac{3 π}{4}$ $+- (3pi)/4$

Explanation:

$cos (\frac{5 π}{4}) = cos (\frac{π}{4} + π) = - \frac{cos π}{4} = - \frac{\sqrt{2}}{2}$ $cos ((5pi)/4) = cos (pi/4 + pi) = - cos pi/4 = -(sqrt2)/2$

$a r c x = arccos (\frac{- \sqrt{2}}{2})$ $arc x = arccos ((-sqrt2)/2)$ ---> $x = \pm \frac{3 π}{4}$ $x = +- (3pi)/4$

Answer 2 · 2015-07-29T15:56:23

$\frac{3 π}{4}$ $(3pi)/4$

Explanation:

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

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

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

$arccos (cos (\frac{5 π}{4}))$ $arccos(cos((5pi)/4))$ is an angle between $0$ $0$ and $π$ $pi$ whose cosine is the same as the cosine of $\frac{5 π}{4}$ $(5pi)/4$ .

The angle we want is $\frac{3 π}{4}$ $(3pi)/4$

We know that $cos (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2}$ $cos((5pi)/4) = -sqrt2/2$
and the Quadrant II angle with cosine equal to $- \frac{\sqrt{2}}{2}$ $-sqrt2/2$
is $\frac{3 π}{4}$ $(3pi)/4$

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How do you evaluate $arccos (cos (5 \frac{π}{4}))$ $arccos(cos(5pi/4))$ ?

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How do you evaluate $arccos (cos (5 \frac{π}{4}))$ $arccos(cos(5pi/4))$ ?