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

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Answer 1 · 2015-07-05T16:53:24

$θ$ $theta$ = 120, 240 degrees

Let $arccos (- \frac{1}{2}) = θ, t h e n cos θ = - \frac{1}{2}$ $arccos(-1/2) =theta, then cos theta= -1/2$

It is known that cos 60 = $\frac{1}{2}$ $1/2$

Since $cos θ$ $cos theta$ is negative in IInd and IVth quadrant, $θ$ $theta$ would be = 180-60 and 180+60, that is 120,240 degrees.

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