How do you evaluate ${cos}^{- 1} [cos (\frac{- 5 π}{3})]$ $cos^(-1) [cos((-5pi)/3)]$ ?

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Answer 1 · 2017-07-21T11:15:41

$\frac{- 5 π}{3}$ $(-5pi)/3$

${cos}^{- 1} [cos (\frac{- 5 π}{3})]$ $cos^-1[cos( (-5pi)/3)]$

Let ${cos}^{- 1} [cos (\frac{- 5 π}{3})] = θ$ $cos^-1[cos( (-5pi)/3)] = theta$ , then

$cos θ = cos (\frac{- 5 π}{3})$ $cos theta = cos( (-5pi)/3)$

$:. theta = (-5pi)/3$

$:.cos^-1[cos( (-5pi)/3)] = (-5pi)/3$ [Ans]

Answer 2 · 2017-07-22T05:23:47

$cos^-1(cos((-5pi)/3)) = pi/3$

Here's the trick to this problem: $cos^-1(theta)$ is only defined on the interval $0 le theta le pi$ .

So, what this problem is asking us to do is find the angle between $0$ and $pi$ whose cosine is the same as the cosine of $(-5pi)/3$ .

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Adding $2pi$ to an angle doesn't change any of its trig functions, so we can say that:

$cos((-5pi)/3) = cos((-5pi)/3 + 2pi)$

$cos((-5pi)/3) = cos(pi/3)$

And since $pi/3$ is between $0$ and $pi$ , we can say that:

$cos^-1(cos((-5pi)/3)) = pi/3$

Final Answer

How do you evaluate cos^(-1) [cos((-5pi)/3)]cos−1[cos(−5π3)]cos^(-1) [cos((-5pi)/3)]?