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Question: How do you evaluate \(\sin \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)\)?...

How do you evaluate sin(cos1(22))\sin \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)?

Explanation

Solution

In the problem we have an inverse trigonometric function and a normal trigonometric function. So, we will first solve the inverse trigonometric function, for this we will assume the inverse trigonometric ratio to a variable let us say xx. Now we will simplify and calculate the value of the inverse trigonometric function. From this value we will calculate the value of the given function by substituting the calculated inverse trigonometric function.

Complete step-by-step answer:
Given that, sin(cos1(22))\sin \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)
In the above equation we have inverse trigonometric function cos1{{\cos }^{-1}} and the normal trigonometric function sin\sin . Assuming the inverse trigonometric function as a variable let us say xx. Then we will have
cos1(22)=x{{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)=x
Applying the cos\cos function on the both sides of the above equation. Then we will get
cos(cos1(22))=cosx\Rightarrow \cos \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)=\cos x
We know that a function will be cancelled when it is multiplied by its inverse function, then we will have
22=cosx\Rightarrow \dfrac{\sqrt{2}}{2}=\cos x
We have the cos(π4)=22\cos \left( \dfrac{\pi }{4} \right)=\dfrac{\sqrt{2}}{2}, substituting this value in the above equation, then we will get
cos(π4)=cosx\Rightarrow \cos \left( \dfrac{\pi }{4} \right)=\cos x
Applying inverse function on both sides of the above equation and cancelling the terms that are possible in the above equation, then we will get
x=π4\Rightarrow x=\dfrac{\pi }{4}
Hence the value of cos1(22){{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) will be
cos1(22)=π4{{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)=\dfrac{\pi }{4}
Substituting this value in the given equation, then we will get
sin(cos1(22))=sin(π4)\Rightarrow \sin \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)=\sin \left( \dfrac{\pi }{4} \right)
We have the value of sin(π4)=22\sin \left( \dfrac{\pi }{4} \right)=\dfrac{\sqrt{2}}{2}.
Hence sin(cos1(22))=22\sin \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)=\dfrac{\sqrt{2}}{2}.

Note: In this problem we have assumed the inverse trigonometric function to a variable for our convenience. If you are familiar with the values of inverse trigonometric functions, then you can simplify the use of that value. But it is not preferable to go with this method. So, follow the above method to evaluate any kind of equation for different angles.