Question

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$\begin{gathered} \bf \: Find \: the \: principal \: value \: of \: \cos^{ - 1} ( \frac{ - 1}{2} ) \\ \end{gathered}$
_______________________________ Class -12
Ch - 2 Inverse Trigonometric Functions​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Method 1

$\bf \: let \: y = \cos^{ - 1} ( \frac{ - 1}{2} )$

We know that

$\bf \cos^{ - 1} (- x) = \pi - \cos^{ - 1} (x)$

$\sf \: y = \pi - cos^{ - 1} ( \frac{1}{2} ) \\ \sf \: y = \pi - \frac{\pi}{3} \\ \sf \: y = \frac{2x}{3} \\\bf Since \: Range \: of cos^{-1 }is [0, \pi] \\ \bf \red{Hence, \: the \: principal \: value \: is \: \frac{2x}{3}}$

Method 2

$\bf \: Let \: y = cos^{ - 1} \frac{ - 1}{2}$

$\sf \cos \: y = \frac{ - 1}{2} \\ \sf \: \cos \: y = \cos \: ( \frac{2\pi}{3} ) \\ \bf Since \: Range \: of cos^{-1 }is [0, \pi] \\ \bf \red{Hence, \: the \: principal \: value \: is \: \frac{2x}{3}}$