How do you evaluate $arccos (- \frac{1}{\sqrt{2}})$ $arccos(-1/sqrt2)$ ?

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How do you evaluate $arccos (- \frac{1}{\sqrt{2}})$ $arccos(-1/sqrt2)$ ?

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Answer 1 · 2016-04-15T02:59:30

As the range is not specified, the general value is given. The principal value of $a r c cos (- \frac{1}{\sqrt{2}})$ $arc cos(-1/sqrt2)$ is $\frac{3 π}{4}$ $(3pi)/4$ .
The general value is $2 k π \pm \frac{3 π}{4}$ $2kpi+-(3pi)/4$ , k = 0. 1. 2. 3.,...

Explanation:

Cosine is positive in the 1st quadrant [0, pi/2] and negative in the second $[\frac{π}{2}, π]$ $[pi/2, pi]$

The principal value for arc cos is defined to be in $[0, π]$ $[0, pi]$ .

The general value is $2 k π \pm$ $2kpi+-$ (principal value), k = 0. 1. 2. 3.,...

Here, the principal value of $a r c cos (- \frac{1}{\sqrt{2}}) = \frac{3 π}{4}$ $arc cos (-1/sqrt2) = (3pi)/4$ .

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How do you evaluate $arccos (- \frac{1}{\sqrt{2}})$ $arccos(-1/sqrt2)$ ?

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How do you evaluate $arccos (- \frac{1}{\sqrt{2}})$ $arccos(-1/sqrt2)$ ?