
How do you calculate the \[sin\left( {si{n^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)\] ?
Answer
542.1k+ views
Hint: For solving this particular question we need not to use any mathematical operations or any complex algorithm. We just have to use a simple property that is \[sin(si{n^{ - 1}}(x)) = x\] , functions sine and inverse sine undo each one another. In order to solve and simplify the given expression we have to use the identity and express our given expression in the simplest form and thereby solve it..
Complete step-by-step solution:
The given expression is ,
\[sin\left( {si{n^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)\]
We have to calculate the value of the above expression.
For solving this there exist a property that is ,
\[sin(si{n^{ - 1}}(x)) = x\] , here functions sine and inverse sine undo each one another.
Therefore, we can apply this and easily get the result as follow ,
\[ \Rightarrow sin\left( {si{n^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) = \dfrac{1}{3}\]
Hence we get the required result.
Note: If we have questions similar in nature as that of above can be approached in a similar manner and we can solve it easily and can find the corresponding result. We just have to use a simple property that is \[sin(si{n^{ - 1}}(x)) = x\] , functions sine and inverse sine undo each one another.
Complete step-by-step solution:
The given expression is ,
\[sin\left( {si{n^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)\]
We have to calculate the value of the above expression.
For solving this there exist a property that is ,
\[sin(si{n^{ - 1}}(x)) = x\] , here functions sine and inverse sine undo each one another.
Therefore, we can apply this and easily get the result as follow ,
\[ \Rightarrow sin\left( {si{n^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) = \dfrac{1}{3}\]
Hence we get the required result.
Note: If we have questions similar in nature as that of above can be approached in a similar manner and we can solve it easily and can find the corresponding result. We just have to use a simple property that is \[sin(si{n^{ - 1}}(x)) = x\] , functions sine and inverse sine undo each one another.
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