How do you find the exact value of ${cos}^{- 1} (- \frac{\sqrt{2}}{2})$ $cos^-1 (-sqrt2/2)$ ?

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How do you find the exact value of ${cos}^{- 1} (- \frac{\sqrt{2}}{2})$ $cos^-1 (-sqrt2/2)$ ?

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Answer 1 · 2015-06-28T22:45:58

${cos}^{- 1} (- \frac{\sqrt{2}}{2}) = 135^{o} or 225^{o}$ $cos^(-1)(-sqrt(2)/2) = 135^o or 225^o$
$XXXX$ $color(white)("XXXX")$ if we restrict the range to $[0, 360^{o})$ $[0,360^o)$

Explanation:

$(- \frac{\sqrt{2}}{2}) = (- \frac{1}{\sqrt{2}})$ $(-sqrt(2)/2) = (-1/sqrt(2))$

If $cos (\frac{1}{\sqrt{2}}) = θ$ $cos(1/sqrt(2)) = theta$ then $cos (θ) = (- \frac{1}{\sqrt{2}})$ $cos(theta) = (-1/sqrt(2))$

If $θ ε [0, 2 π)$ $theta epsilon [0,2pi)$
we have the two possibilities indicated in the diagram below:
enter image source here
with $θ$ $theta$ (measuring from the positive x-axis) as either
$180^{o} - 45^{o} = 135^{o}$ $180^o - 45^o = 135^o$
or
$180^{o} + 45^{o} = 225^{o}$ $180^o + 45^o = 225^o$

For people who prefer their angles in radians that is
$π - \frac{π}{4} = \frac{3 π}{4}$ $pi - pi/4 = (3pi)/4$
or
$π + \frac{π}{4} = \frac{5 π}{4}$ $pi+pi/4 = (5pi)/4$

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How do you find the exact value of ${cos}^{- 1} (- \frac{\sqrt{2}}{2})$ $cos^-1 (-sqrt2/2)$ ?

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How do you find the exact value of ${cos}^{- 1} (- \frac{\sqrt{2}}{2})$ $cos^-1 (-sqrt2/2)$ ?