
Find the value of \[{\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)\].
A. 0
B. \[\dfrac{\pi }{2}\]
C. \[\dfrac{{2\pi }}{3}\]
D. \[\dfrac{{10\pi }}{3}\]
Answer
161.7k+ views
First we will simplify the \[\dfrac{{5\pi }}{3}\]. Then we will apply the inverse formula to calculate the value of the given expression.
Formula used:
\[\cos \left( {2\pi - \theta } \right) = \cos \theta \]
\[\sin \left( {2\pi - \theta } \right) = - \sin \theta \]
\[{\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta \] when \[x\in \left[0,\pi\right]\]
\[{\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta \] when \[x\in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\]
Complete step by step solution:
Given expression is \[{\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)\].
We will simplify \[\dfrac{{5\pi }}{3}\]. Rewrite \[\dfrac{{5\pi }}{3}\] as \[2\pi - \dfrac{\pi }{3}\].
\[ = {\cos ^{ - 1}}\left( {\cos \left( {2\pi - \dfrac{\pi }{3}} \right)} \right) + {\sin ^{ - 1}}\left( {\sin \left( {2\pi - \dfrac{\pi }{3}} \right)} \right)\]
Now trigonometry of angle formula
\[ = {\cos ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right) + {\sin ^{ - 1}}\left( { - \sin \dfrac{\pi }{3}} \right)\]
Now we will rewrite \[ - \sin \dfrac{\pi }{3}\]
\[ = {\cos ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right) + {\sin ^{ - 1}}\left( {\sin \left( { - \dfrac{\pi }{3}} \right)} \right)\]
Apply the inverse formula
\[ = \dfrac{\pi }{3} + \left( { - \dfrac{\pi }{3}} \right)\]
Simplify the above expression
\[ = \dfrac{\pi }{3} - \dfrac{\pi }{3}\]
\[ = 0\]
Hence option A is the correct option.
Additional information: In the first quadrant all trigonometry ratios are positive. In the second quadrant \[\sin\] and \[\csc\] functions are positive. In the third quadrant \[\tan\] and \[\cot\] functions are positive. In the fourth quadrant \[\cos\] and \[\sec\] functions are positive.
\[\dfrac{{5\pi }}{{3}}\] lies on the fourth quadrant. So only \[\cos\] is positive and \[\sin\] is negative.
Note: Students often do a common mistake to find the value of \[{\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)\]. They directly apply the inverse formulas \[{\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta \] and \[{\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta \]. They get \[{\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right) = \dfrac{{5\pi }}{3} + \dfrac{{5\pi }}{3} = \dfrac{{10\pi }}{3}\] which wrong answer. The correct answer is 0.
Formula used:
\[\cos \left( {2\pi - \theta } \right) = \cos \theta \]
\[\sin \left( {2\pi - \theta } \right) = - \sin \theta \]
\[{\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta \] when \[x\in \left[0,\pi\right]\]
\[{\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta \] when \[x\in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\]
Complete step by step solution:
Given expression is \[{\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)\].
We will simplify \[\dfrac{{5\pi }}{3}\]. Rewrite \[\dfrac{{5\pi }}{3}\] as \[2\pi - \dfrac{\pi }{3}\].
\[ = {\cos ^{ - 1}}\left( {\cos \left( {2\pi - \dfrac{\pi }{3}} \right)} \right) + {\sin ^{ - 1}}\left( {\sin \left( {2\pi - \dfrac{\pi }{3}} \right)} \right)\]
Now trigonometry of angle formula
\[ = {\cos ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right) + {\sin ^{ - 1}}\left( { - \sin \dfrac{\pi }{3}} \right)\]
Now we will rewrite \[ - \sin \dfrac{\pi }{3}\]
\[ = {\cos ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right) + {\sin ^{ - 1}}\left( {\sin \left( { - \dfrac{\pi }{3}} \right)} \right)\]
Apply the inverse formula
\[ = \dfrac{\pi }{3} + \left( { - \dfrac{\pi }{3}} \right)\]
Simplify the above expression
\[ = \dfrac{\pi }{3} - \dfrac{\pi }{3}\]
\[ = 0\]
Hence option A is the correct option.
Additional information: In the first quadrant all trigonometry ratios are positive. In the second quadrant \[\sin\] and \[\csc\] functions are positive. In the third quadrant \[\tan\] and \[\cot\] functions are positive. In the fourth quadrant \[\cos\] and \[\sec\] functions are positive.
\[\dfrac{{5\pi }}{{3}}\] lies on the fourth quadrant. So only \[\cos\] is positive and \[\sin\] is negative.
Note: Students often do a common mistake to find the value of \[{\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right)\]. They directly apply the inverse formulas \[{\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta \] and \[{\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta \]. They get \[{\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{3}} \right) = \dfrac{{5\pi }}{3} + \dfrac{{5\pi }}{3} = \dfrac{{10\pi }}{3}\] which wrong answer. The correct answer is 0.
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