Question

Find the principal value of cos^-1(-1/√3)
with steps

Answer 1

Answer:

I guess your question would be,

$\sf cot^{-1} (\frac{-1}{\sqrt{3}})$

As, $\sf \cos \theta$ $\sf{ don't\;have} \frac{1}{\sqrt{3}}$

$\sf{Let}$,

$\sf cot^{-1} (\frac{-1}{\sqrt{3}}) = \sf{y}$

$\sf \cot y = \frac{-1}{\sqrt{3}}$

$\sf \cot y = - \cot (\frac{π}{\sqrt{3}})$

$\sf \cot y = \cot (π-\frac{π}{\sqrt{3}})$

$\sf \cot y = \cot (\frac{2π}{3})$

$\sf{y} = \frac{2π}{3}$

We know that the range of $\sf cot^{-1}$ is (0,π) , so you can see that the y which is obtained is between the range

Hence principal value of $\sf cot^{-1} (\frac{-1}{\sqrt{3}}) = \frac{2π}{3}$

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Answer 2

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