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

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

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Answer 1 · 2018-05-05T06:36:38

Refer to Explanation

Explanation:

${cos}^{- 1} (\frac{1}{2}) = 60^{\circ}$ $cos^-1(1/2) = 60^@$

Trigonometric Ratios of Special Angles:
![https://www.basic-mathematics.com/http://trigonometric-ratios-of-special-angles.html](https://useruploads.socratic.org/CWElR8lRtm5IxQlgh4CQ_trigonometric-values-of-special-angles.png)

Answer 2 · 2018-05-05T16:51:12

$\frac{π}{3}; \frac{5 π}{3}$ $pi/3; (5pi)/3$
$60^{\circ}; 300^{\circ}$ $60^@; 300^@$

Explanation:

$cos x = \frac{1}{2}$ $cos x= 1/2$
Trig table gives -->
$x = \frac{π}{3}$ $x = pi/3$ , or $x = 60^{\circ}$ $x = 60^@$
Unit circle gives another arc (angle) that has the same cos value (1/2) -->
$x = - \frac{π}{3}$ $x = - pi/3$ , or $x = - 60^{\circ}$ $x = - 60^@$ ,
or $x = \frac{5 π}{3}$ $x = (5pi)/3$ (co-terminal to $- \frac{π}{3}$ $- pi/3$ )

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

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