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Question: How do you find the exact value of \[\sin \left( {{{\sin }^{ - 1}}\left( {0.3} \right)} \right)\] ?...

How do you find the exact value of sin⁡(sin⁡−1(0.3))\sin \left( {{{\sin }^{ - 1}}\left( {0.3} \right)} \right) ?

Explanation

Solution

Hint : We will use the question itself to solve the question. We will use substitution here. The inverse function will be submitted as a variable. Then we will take the sine function value on both the sides. That will directly give the answer to us.

Complete step-by-step answer :
Given that,
sin⁡(sin⁡−1(0.3))\sin \left( {{{\sin }^{ - 1}}\left( {0.3} \right)} \right)
Let,
sin⁡−1(0.3)=α{\sin ^{ - 1}}\left( {0.3} \right) = \alpha
Then taking sin function on both the sides,
0.3=sin⁡α0.3 = \sin \alpha
Now we can clearly see that,
sin⁡(sin⁡−1(0.3))=sin⁡α\sin \left( {{{\sin }^{ - 1}}\left( {0.3} \right)} \right) = \sin \alpha
And value of sin⁡α\sin \alpha is 0.3. so we conclude that,
sin⁡(sin⁡−1(0.3))=0.3\sin \left( {{{\sin }^{ - 1}}\left( {0.3} \right)} \right) = 0.3
This is our final answer.
So, the correct answer is “0.3”.

Note : Here note that we can directly find the answer as sin⁡(sin⁡−1θ)=θ\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta but the solution needs to be elaborated. If sometimes the combination of function is used then we need to find the solution the way we found here.
Sometimes the inverse functions are written as arc functions like
sin⁡−1θ⇒arcsin⁡θ{\sin ^{ - 1}}\theta \Rightarrow \arcsin \theta . So do read the question first and don’t get confused. There are some identities in trigonometry related to this inverse functionality.