Question

How do you evaluate \[\sin \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)\] ?

Answer 1

Hint: This is a pretty straight forward question and requires some basic knowledge of trigonometry. We also need to know about the domain and range of the trigonometric functions as well as their inverse functions. The function \[\sin \left( {{\sin }^{-1}}\left( x \right) \right)\] will evaluate to any value between \[\left[ -1,1 \right]\] . The plot of the graph is same as that of \[y=x,x\in \left[ -1,1 \right]\] .

Complete step by step answer:
The domain and ranges of the following trigonometric functions are as follows,

Function	Domain	Range
\[f\left( x \right)=\sin x\]	\[\left( -\infty ,+\infty \right)\]	\[\left[ -1,1 \right]\]
\[f\left( x \right)=\cos x\]	\[\left( -\infty ,+\infty \right)\]	\[\left[ -1,1 \right]\]
\[f\left( x \right)=\tan x\]	– all real numbers except \[\left( \dfrac{\pi }{2}+n\pi \right)\]	\[\left( -\infty ,+\infty \right)\]
\[f\left( x \right)=\operatorname{cosec}x\]	all real numbers except \[\left( n\pi \right)\]	\[(-\infty ,-1]\bigcup{[1,+\infty )}\]
\[f\left( x \right)=\sec x\]	all real numbers except \[\left( \dfrac{\pi }{2}+n\pi \right)\]	\[(-\infty ,-1]\bigcup{[1,+\infty )}\]
\[f\left( x \right)=\cot x\]	- all real numbers except \[\left( n\pi \right)\]	\[\left( -\infty ,+\infty \right)\]
\[f\left( x \right)={{\sin }^{-1}}x\]	\[\left[ -1,1 \right]\]	\[\left[ -\dfrac{\pi }{2},+\dfrac{\pi }{2} \right]\]
\[f\left( x \right)={{\cos }^{-1}}x\]	\[\left[ -1,1 \right]\]	\[\left[ 0,\pi \right]\]
\[f\left( x \right)={{\tan }^{-1}}x\]	\[\left( -\infty ,+\infty \right)\]	\[\left[ -\dfrac{\pi }{2},+\dfrac{\pi }{2} \right]\]
\[f\left( x \right)={{\operatorname{cosec}}^{-1}}x\]	\[\left( -\infty ,-1 \right]\bigcup{\left[ 1,+\infty \right)}\]	\[\left[ -\pi ,-\dfrac{\pi }{2} \right]\bigcup{\left[ 0,\dfrac{\pi }{2} \right]}\]
\[f\left( x \right)={{\sec }^{-1}}x\]	\[\left( -\infty ,-1 \right]\bigcup{\left[ 1,+\infty \right)}\]	\[\left[ 0,\dfrac{\pi }{2} \right]\bigcup{\left[ \pi ,\dfrac{3\pi }{2} \right]}\]
\[f\left( x \right)={{\cot }^{-1}}x\]	\[\left( -\infty ,+\infty \right)\]	\[\left[ 0,\pi \right]\]

For \[y={{\sin }^{-1}}x\] in the domain \[\left[ -1,1 \right]\] we can write \[x=\sin y\]. Since in this problem \[\dfrac{3}{5}\] lies in the domain of \[{{\sin }^{-1}}x\] , we can evaluate is very easily.

Now, starting off with our solution, we can write from the given problem that,
Let \[y={{\sin }^{-1}}\dfrac{3}{5}\]
\[\Rightarrow \sin y=\dfrac{3}{5}\]
Now, according to the problem statement,
\[\sin \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)\]
We can evaluate it as,
\[\begin{align}
  & \sin \left( y \right) \\
 & \Rightarrow \dfrac{3}{5} \\
\end{align}\]

Since previously we have found out that \[\sin y=\dfrac{3}{5}\] . Thus the answer to this problem should be \[\dfrac{3}{5}\] .

Note: In these types of questions we need to be very careful about the domain and range of the given trigonometric function. If the value given inside of an inverse function is not within the domain then we need to make necessary operations following the graph of the inverse trigonometric function. An easy way to do such problems is to remember some of the values like, \[\sin {{0}^{\circ }}=0,\sin {{37}^{\circ }}=\dfrac{3}{5},\sin {{53}^{\circ }}=\dfrac{4}{5},\sin {{90}^{\circ }}=1\] and putting the values at times of need. The problem can also be done graphically by plotting it in a graph paper and analysing it.