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Question: The value of \[{\sin ^{ - 1}}\left( {\dfrac{{2\sqrt 2 }}{3}} \right) + {\sin ^{ - 1}}\left( {\dfrac{...

The value of sin⁡−1(223)+sin⁡−1(13){\sin ^{ - 1}}\left( {\dfrac{{2\sqrt 2 }}{3}} \right) + {\sin ^{ - 1}}\left( {\dfrac{1}{3}} \right) is:
A). π6\dfrac{\pi }{6}
B). π4\dfrac{\pi }{4}
C). π2\dfrac{\pi }{2}
D). 2π3\dfrac{{2\pi }}{3}
E). 00

Explanation

Solution

In the given question, we have been given an expression involving sum of inverse of two trigonometric expressions. In the two expressions, the functions are the same, but the arguments are different. We have to simplify and find the value of the expression. To do that, we are going to convert the expressions into such forms that their arguments become equal, it does not matter if we have to change the functions, so long as the arguments are becoming equal.

Formula used:
We are going to use the formula of sum of inverse of sine and cosine:
sin⁡−1A+cos⁡−1A=π2{\sin ^{ - 1}}A + {\cos ^{ - 1}}A = \dfrac{\pi }{2}

Complete step by step solution:
We have to solve p=sin⁡−1(223)+sin⁡−1(13)p = {\sin ^{ - 1}}\left( {\dfrac{{2\sqrt 2 }}{3}} \right) + {\sin ^{ - 1}}\left( {\dfrac{1}{3}} \right).
Let k=sin⁡−1(223)k = {\sin ^{ - 1}}\left( {\dfrac{{2\sqrt 2 }}{3}} \right), then,
sin⁡k=223\sin k = \dfrac{{2\sqrt 2 }}{3}
Now, we know, sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
So, we have,
sin⁡2k+cos⁡2k=1{\sin ^2}k + {\cos ^2}k = 1
Putting in the value,
89+cos⁡2k=1⇒cos⁡2k=19\dfrac{8}{9} + {\cos ^2}k = 1 \Rightarrow {\cos ^2}k = \dfrac{1}{9}
So, cos⁡k=13⇒k=cos⁡−1(13)\cos k = \dfrac{1}{3} \Rightarrow k = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)
Hence, we get,
p=cos⁡−1(13)+sin⁡−1(13)p = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right) + {\sin ^{ - 1}}\left( {\dfrac{1}{3}} \right)
Using the formula of sum of inverse of sine and cosine:
sin⁡−1A+cos⁡−1A=π2{\sin ^{ - 1}}A + {\cos ^{ - 1}}A = \dfrac{\pi }{2}
p=π2p = \dfrac{\pi }{2}
Hence, the correct option is C.

Note: In the given question, we had to find the value of an expression in which we were given two trigonometric expressions involving the sum of inverses of two equal functions but different arguments. We had to solve that. We did it by making the arguments equal by using the appropriate identities and formulae. It is thus very important that we know all the basics of the concept of inverses and know how to transform one expression to another.