How do you evaluate $a r c cos (- \frac{1}{\sqrt{2}})$ $arc cos (-1/sqrt2)$ ?

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How do you evaluate $a r c cos (- \frac{1}{\sqrt{2}})$ $arc cos (-1/sqrt2)$ ?

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Answer 1 · 2016-12-15T14:41:43

$\frac{3 π}{4}; \frac{5 π}{4}$ $(3pi)/4; (5pi)/4$

Explanation:

Use trig table of special arcs and unit circle.
$cos x = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$ $cos x = - 1/sqrt2 = - sqrt2/2$ -->
arc $x = \pm \frac{3 π}{4}$ $x = +- (3pi)/4$
The arc $x = - \frac{3 π}{4}$ $x = - (3pi)/4$ is co-terminal to $x = \frac{5 π}{4}$ $x = (5pi)/4$
There for, the answers for $(0, 2 π)$ $(0, 2pi)$ are:
$\frac{3 π}{4}; \frac{5 π}{4}$ $(3pi)/4; (5pi)/4$

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How do you evaluate $a r c cos (- \frac{1}{\sqrt{2}})$ $arc cos (-1/sqrt2)$ ?

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