Answer:
[tex]\boxed{\bf\:{cos}^{ - 1}\bigg( - \dfrac{1}{2} \bigg) = \dfrac{2\pi}{3} \: }\\ [/tex]
Step-by-step explanation:
Given inverse trigonometric function is
[tex]\sf\: {cos}^{ - 1}\bigg( - \dfrac{1}{2} \bigg)\\ [/tex]
We know,
[tex]\boxed{\sf\:{cos}^{ - 1}( - x) = \pi - {cos}^{ - 1}x \: } \\ [/tex]
So, Using this result, we get
[tex]\sf\: = \: \pi - {cos}^{ - 1}\bigg( \dfrac{1}{2} \bigg)\\ [/tex]
[tex]\sf\: = \: \pi - {cos}^{ - 1}\bigg(cos \dfrac{\pi}{3} \bigg)\\ [/tex]
[tex]\boxed{\bf\:{cos}^{ - 1}(cosx) = x \: \: \forall \: x \: \in \: \left[0, \: \pi \right] \: } \\ [/tex]
[tex]\sf\: = \: \pi - \dfrac{\pi}{3} \\ [/tex]
[tex]\sf\: = \: \dfrac{3\pi - \pi}{3} \\ [/tex]
[tex]\sf\: = \: \dfrac{2\pi}{3} \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:{cos}^{ - 1}\bigg( - \dfrac{1}{2} \bigg) = \dfrac{2\pi}{3} \: }\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = {sin}^{ - 1}(sinx) & \sf x \: \: if -\dfrac{\pi }{2} \leqslant x \leqslant \dfrac{\pi }{2}\\ \\ \sf x= {cos}^{ - 1}(cosx) & \sf x \: \: if \: 0 \leqslant y \leqslant \pi \\ \\ \sf y = {tan}^{ - 1}(tanx) & \sf x \: \: if \: - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}\\ \\ \sf y = {cosec}^{ - 1}(cosecx) & \sf x \: \: if \: x \: \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi }{2}\bigg] - \{0 \}\\ \\ \sf y = {sec}^{ - 1}(secx) & \sf x \: \: if \: x \: \in \: [0, \: \pi] \: - \: \bigg\{\dfrac{\pi }{2}\bigg\}\\ \\ \sf y = {cot}^{ - 1}(cotx) & \sf x \: \: if \: \: \in \: \bigg( - \dfrac{\pi }{2} , \dfrac{\pi }{2}\bigg) - \{0 \} \end{array}} \\ \end{gathered} \\[/tex]
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Answers & Comments
Answer:
[tex]\boxed{\bf\:{cos}^{ - 1}\bigg( - \dfrac{1}{2} \bigg) = \dfrac{2\pi}{3} \: }\\ [/tex]
Step-by-step explanation:
Given inverse trigonometric function is
[tex]\sf\: {cos}^{ - 1}\bigg( - \dfrac{1}{2} \bigg)\\ [/tex]
We know,
[tex]\boxed{\sf\:{cos}^{ - 1}( - x) = \pi - {cos}^{ - 1}x \: } \\ [/tex]
So, Using this result, we get
[tex]\sf\: = \: \pi - {cos}^{ - 1}\bigg( \dfrac{1}{2} \bigg)\\ [/tex]
[tex]\sf\: = \: \pi - {cos}^{ - 1}\bigg(cos \dfrac{\pi}{3} \bigg)\\ [/tex]
We know,
[tex]\boxed{\bf\:{cos}^{ - 1}(cosx) = x \: \: \forall \: x \: \in \: \left[0, \: \pi \right] \: } \\ [/tex]
So, Using this result, we get
[tex]\sf\: = \: \pi - \dfrac{\pi}{3} \\ [/tex]
[tex]\sf\: = \: \dfrac{3\pi - \pi}{3} \\ [/tex]
[tex]\sf\: = \: \dfrac{2\pi}{3} \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:{cos}^{ - 1}\bigg( - \dfrac{1}{2} \bigg) = \dfrac{2\pi}{3} \: }\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = {sin}^{ - 1}(sinx) & \sf x \: \: if -\dfrac{\pi }{2} \leqslant x \leqslant \dfrac{\pi }{2}\\ \\ \sf x= {cos}^{ - 1}(cosx) & \sf x \: \: if \: 0 \leqslant y \leqslant \pi \\ \\ \sf y = {tan}^{ - 1}(tanx) & \sf x \: \: if \: - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}\\ \\ \sf y = {cosec}^{ - 1}(cosecx) & \sf x \: \: if \: x \: \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi }{2}\bigg] - \{0 \}\\ \\ \sf y = {sec}^{ - 1}(secx) & \sf x \: \: if \: x \: \in \: [0, \: \pi] \: - \: \bigg\{\dfrac{\pi }{2}\bigg\}\\ \\ \sf y = {cot}^{ - 1}(cotx) & \sf x \: \: if \: \: \in \: \bigg( - \dfrac{\pi }{2} , \dfrac{\pi }{2}\bigg) - \{0 \} \end{array}} \\ \end{gathered} \\[/tex]