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Question: How do you find the exact value of \[{{\sin }^{-1}}\left( \cos \left( 2\dfrac{\pi }{3} \right) \righ...

How do you find the exact value of sin⁡−1(cos⁡(2π3)){{\sin }^{-1}}\left( \cos \left( 2\dfrac{\pi }{3} \right) \right)?

Explanation

Solution

In the given function, we have been asked to find the exact value of given trigonometric expression. In order to find the value, first we need to simplify the given expression in a way we can apply trigonometric identity. Then by using difference identity of cosine function i.e. cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b, we will expand and further using the trigonometric ratios table simplify the expression further and we will get the exact value of the given expression.

Complete step by step solution:
We have given that,
sin⁡−1(cos⁡(2π3)){{\sin }^{-1}}\left( \cos \left( 2\dfrac{\pi }{3} \right) \right)
Rewrite the above expression as,
sin⁡−1(cos⁡(π−π3)){{\sin }^{-1}}\left( \cos \left( \pi -\dfrac{\pi }{3} \right) \right)
Using the difference identity of cosine, i.e. cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b.
Expanding the above expression, we obtained
sin⁡−1(cos⁡(π−π3))=sin⁡−1(cos⁡πcos⁡π3+sin⁡πsin⁡π3){{\sin }^{-1}}\left( \cos \left( \pi -\dfrac{\pi }{3} \right) \right)={{\sin }^{-1}}\left( \cos \pi \cos \dfrac{\pi }{3}+\sin \pi \sin \dfrac{\pi }{3} \right)
Using the trigonometric ratios table;
We know that the value of,
cos⁡π=−1\cos \pi =-1
cos⁡π3=12\cos \dfrac{\pi }{3}=\dfrac{1}{2}
sin⁡π=0\sin \pi =0
sin⁡π3=32\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}
Applying these values in the above expressions, we get
sin⁡−1(cos⁡(π−π3))=sin⁡−1(−1×12+0×32)=sin⁡−1(−12){{\sin }^{-1}}\left( \cos \left( \pi -\dfrac{\pi }{3} \right) \right)={{\sin }^{-1}}\left( -1\times \dfrac{1}{2}+0\times \dfrac{\sqrt{3}}{2} \right)={{\sin }^{-1}}\left( -\dfrac{1}{2} \right)
Now,
Let u=sin⁡−1(−12)u={{\sin }^{-1}}\left( -\dfrac{1}{2} \right)
Then, we have
sin⁡u=−12\sin u=\dfrac{-1}{2}
Since the sine function is negative in third and fourth quadrants;
Therefore, u < 2π2\pi
⇒sin⁡(π+π6)=−sin⁡π6 ⇒ u=π+π6=7π6\Rightarrow \sin \left( \pi +\dfrac{\pi }{6} \right)=-\sin \dfrac{\pi }{6}\ \Rightarrow \ u=\pi +\dfrac{\pi }{6}=\dfrac{7\pi }{6}
And
⇒sin⁡(2π−π6)=−sin⁡π6 ⇒ u=2π−π6=11π6\Rightarrow \sin \left( 2\pi -\dfrac{\pi }{6} \right)=-\sin \dfrac{\pi }{6}\ \Rightarrow \ u=2\pi -\dfrac{\pi }{6}=\dfrac{11\pi }{6}
Therefore,
The exact values of sin⁡−1(cos⁡(2π3)){{\sin }^{-1}}\left( \cos \left( 2\dfrac{\pi }{3} \right) \right) are 7π6\dfrac{7\pi }{6} and 11π6\dfrac{11\pi }{6}.
Hence, it is the required answer.

Note: In order to solve these types of questions, you should always need to remember the properties of trigonometric and the trigonometric ratios as well. It will make questions easier to solve. It is preferred that while solving these types of questions we should carefully examine the pattern of the given function and then you would apply the formulas according to the pattern observed. As if you directly apply the formula it will create confusion ahead and we will get the wrong answer.