How do you solve and find the value of $ \sin \left( {{\sin }^{-1}}\left( \dfrac{3}{4} \right) \right) $ ?
Answer
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Hint: We know that if the value of sin x is equal to y then we can write $ x={{\sin }^{-1}}y $ where y is in between -1 to 1. We can replace x as $ {{\sin }^{-1}}y $ in $ y=\sin x $ because we have written the formula $ x={{\sin }^{-1}}y $ , so replacing x as $ {{\sin }^{-1}}y $ in $ y=\sin x $ we get $ y=\sin \left( {{\sin }^{-1}}y \right) $ . we can use this formula to find the value of $ \sin \left( {{\sin }^{-1}}\left( \dfrac{3}{4} \right) \right) $
Complete step by step answer:
We have evaluate the value of $ \sin \left( {{\sin }^{-1}}\left( \dfrac{3}{4} \right) \right) $
Let’s assume the value of $ {{\sin }^{-1}}\left( \dfrac{3}{4} \right) $ is equal to x, so we can write $ x={{\sin }^{-1}}\left( \dfrac{3}{4} \right) $ ….eq1
Then $ \dfrac{3}{4} $ is equal to sin x, we can write $ \sin x=\dfrac{3}{4} $ ……eq2
Now we can replace x by $ {{\sin }^{-1}}\left( \dfrac{3}{4} \right) $ in eq2 , we can see in eq1 $ {{\sin }^{-1}}\left( \dfrac{3}{4} \right) $ is equal to x
So by replacing we get
$ \sin \left( {{\sin }^{-1}}\left( \dfrac{3}{4} \right) \right)=\dfrac{3}{4} $
So the value of $ \sin \left( {{\sin }^{-1}}\left( \dfrac{3}{4} \right) \right) $ is equal to $ \dfrac{3}{4} $.
Note:
We can remember $ y=\sin \left( {{\sin }^{-1}}y \right) $ as a standard formula, we don’t have to solve all this. But the reverse is not always true $ {{\sin }^{-1}}\left( \sin y \right) $ is not always equal to y. $ {{\sin }^{-1}}\left( \sin y \right) $ is equal to y when y lies in between $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ because the range of $ {{\sin }^{-1}}x $ is $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ . If the value of y in $ {{\sin }^{-1}}\left( \sin y \right) $ is not lie in the range $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ then we have to find a number x in the range $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ such that sin x = sin y then the value of $ {{\sin }^{-1}}\left( \sin y \right) $ will equal to x. Always remember that the value of $ {{\sin }^{-1}}x $ will exist when x is from -1 to 1. Beyond that $ {{\sin }^{-1}}x $ does not exist.
Complete step by step answer:
We have evaluate the value of $ \sin \left( {{\sin }^{-1}}\left( \dfrac{3}{4} \right) \right) $
Let’s assume the value of $ {{\sin }^{-1}}\left( \dfrac{3}{4} \right) $ is equal to x, so we can write $ x={{\sin }^{-1}}\left( \dfrac{3}{4} \right) $ ….eq1
Then $ \dfrac{3}{4} $ is equal to sin x, we can write $ \sin x=\dfrac{3}{4} $ ……eq2
Now we can replace x by $ {{\sin }^{-1}}\left( \dfrac{3}{4} \right) $ in eq2 , we can see in eq1 $ {{\sin }^{-1}}\left( \dfrac{3}{4} \right) $ is equal to x
So by replacing we get
$ \sin \left( {{\sin }^{-1}}\left( \dfrac{3}{4} \right) \right)=\dfrac{3}{4} $
So the value of $ \sin \left( {{\sin }^{-1}}\left( \dfrac{3}{4} \right) \right) $ is equal to $ \dfrac{3}{4} $.
Note:
We can remember $ y=\sin \left( {{\sin }^{-1}}y \right) $ as a standard formula, we don’t have to solve all this. But the reverse is not always true $ {{\sin }^{-1}}\left( \sin y \right) $ is not always equal to y. $ {{\sin }^{-1}}\left( \sin y \right) $ is equal to y when y lies in between $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ because the range of $ {{\sin }^{-1}}x $ is $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ . If the value of y in $ {{\sin }^{-1}}\left( \sin y \right) $ is not lie in the range $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ then we have to find a number x in the range $ -\dfrac{\pi }{2} $ to $ \dfrac{\pi }{2} $ such that sin x = sin y then the value of $ {{\sin }^{-1}}\left( \sin y \right) $ will equal to x. Always remember that the value of $ {{\sin }^{-1}}x $ will exist when x is from -1 to 1. Beyond that $ {{\sin }^{-1}}x $ does not exist.
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