Question

How do you find the value of \[\sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right]?\]

Answer 1

Hint: We are asked to find the value of \[\sin \left( 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right),\] then we start our solution by considering \[{{\cos }^{-1}}\left( \dfrac{3}{5} \right)\] as \[\theta \] and then after that we use \[\sin \theta =2\sin \theta \cos \theta .\] After this we learn about how sin and sin inverse are connected then later use their relationships to solve further and we will also use \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to change to \[\cos \theta \] into \[\sin \theta \] using all this we simplify and solve our problem.

Complete step by step answer:
We are given a function as \[\sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right],\] we have to evaluate the value of this function, to do so we will learn that how trigonometric function is related to their inverses then we also convert our trigonometric function to other. Now, we have given that \[\sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right].\] We consider \[{{\cos }^{-1}}\left( \dfrac{3}{5} \right)\] as \[\theta \] and then we get our equation as
\[\sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right]=\sin 2\theta \]
As we know that, \[\sin 2\theta =2\sin \theta \cos \theta ,\] so we get,
\[\Rightarrow \sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right]=2\sin \left[ {{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right].\cos \left[ {{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right]......\left( i \right)\]
Now to simplify it further, we will learn how the ratio is connected to its inverse. For sin, we have that \[\sin \left( {{\sin }^{-1}}\theta \right)=\theta \] and also for cos, we have \[\cos \left( {{\cos }^{-1}}\theta \right)=\theta .\] So, we apply there on \[\cos \left( {{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right),\] we get that \[\cos \left( {{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right)=\dfrac{3}{5}.......\left( ii \right)\]
And to simplify \[\sin \left( {{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right),\] we will change \[{{\cos }^{-1}}\left( \dfrac{3}{5} \right)\] to \[{{\sin }^{-1}}\] and then only we can reduce it further. Now, as we considered earlier, \[\theta ={{\cos }^{-1}}\left( \dfrac{3}{5} \right)\] so \[\cos \theta =\dfrac{3}{5}.\]
As we know \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,\] so we get,
\[\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]
\[\Rightarrow \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }\]
Using the above values, we get,
\[\Rightarrow \sin \theta =\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}\]
On simplifying, we get,
\[\Rightarrow \sin \theta =\sqrt{\dfrac{25-9}{25}}\]
\[\Rightarrow \sin \theta =\dfrac{4}{5}\]
\[\Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{4}{5} \right)\]
Here, we get, \[\theta ={{\sin }^{-1}}\left( \dfrac{4}{5} \right)\] is same as \[\theta ={{\sin }^{-1}}\left( \dfrac{4}{5} \right)......\left( iii \right)\]
So, using (ii) and (iii) in (i), we get,
\[\sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right]=2\sin \left( {{\sin }^{-1}}\left( \dfrac{4}{5} \right) \right)\cos \left( {{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right)\]
\[\Rightarrow \sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right]=2\times \dfrac{4}{5}\times \dfrac{3}{5}\]
\[\Rightarrow \sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right]=\dfrac{24}{25}\]
So, we get that our required value of \[\sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right]\] is \[\dfrac{24}{25}.\]

Note:
While solving such problems we should be very careful with the identity like \[\sin 2x\ne 2\sin x\] or \[\cos 2\theta \ne 2\cos \theta \sin \theta .\] We should not mix or use inappropriate identity. Also, we should always cross-check the solution so that the chance of error will get eliminated. While solving the fraction we always report the answer in the simplest form. So, if something is common, we need to cancel it always.