Question

How do you evaluate \[{{\sin }^{-1}}\left( \sin \left( \dfrac{4\pi }{3} \right) \right)\] without a calculator?

Answer 1

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

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 \[\left( \dfrac{\pi }{2}+n\pi \right)\]	\[\left( -\infty ,+\infty \right)\]
\[f\left( x \right)=\text{cosec}x\]	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 \[\left( n\pi \right)\]	\[\left( -\infty ,+\infty \right)\]
\[f\left( x \right)={{\sin }^{-1}}x\]	\[\left[ -1,1 \right]\]	\[\left[ -\dfrac{\pi }{2},+\dfrac{\pi }{2} \right]\]
\[f\left( x \right)={{\cos }^{-1}}x\]	\[\left[ -1,1 \right]\]	\[\left[ 0,\pi \right]\]
\[f\left( x \right)={{\tan }^{-1}}x\]	\[\left( -\infty ,+\infty \right)\]	\[\left[ -\dfrac{\pi }{2},+\dfrac{\pi }{2} \right]\]
\[f\left( x \right)=\text{cose}{{\text{c}}^{-1}}x\]	\[\left( -\infty ,-1 \right]\bigcup{\left[ 1,+\infty \right)}\]	\[\left[ -\pi ,-\dfrac{\pi }{2} \right]\bigcup{\left[ 0,\dfrac{\pi }{2} \right]}\]
\[f\left( x \right)={{\sec }^{-1}}x\]	\[\left( -\infty ,-1 \right]\bigcup{\left[ 1,+\infty \right)}\]	\[\left[ 0,\dfrac{\pi }{2} \right]\bigcup{\left[ \pi ,\dfrac{3\pi }{2} \right]}\]
\[f\left( x \right)={{\cot }^{-1}}x\]	\[\left( -\infty ,+\infty \right)\]	\[\left[ 0,\pi \right]\]

For \[y={{\sin }^{-1}}x\] in the domain \[\left[ -1,1 \right]\] we can write \[x=\sin y\].
In this problem, first of all what we need is to represent \[\dfrac{4\pi }{3}\] in terms of summation of \[\pi \] and some other small terms. We can write is very easily as,
\[\dfrac{4\pi }{3}=\pi +\dfrac{\pi }{3}\]

Now, starting off with our solution, we can write from the given problem that, we need to find the value of,
\[{{\sin }^{-1}}\left( \sin \left( \pi +\dfrac{\pi }{3} \right) \right)\] , which is the same as that of the given problem.
From trigonometry theory, we know that \[\pi +\dfrac{\pi }{3}\] lies in the third quadrant, and in that quadrant, \[\sin x\] has a negative value. Only \[\tan x\] and \[\cot x\] has a positive value in that quadrant.
So, we can further reduce our problem, to (by taking a negative sign common, as the angle lies in the third quadrant),
\[\Rightarrow -{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{3} \right) \right)\]
Now, from the above problem, we can say that,
\[\dfrac{\pi }{3}\] lies in the first quadrant, in other words, \[0\le \dfrac{\pi }{3}\le \pi \] and in this portion, \[\sin x\] has a positive value. So we can safely write,
\[\Rightarrow {{\sin }^{-1}}\left( \sin \left( x \right) \right)=\sin \left( x \right)\]
Thus, for our given problem we get,
\[\Rightarrow -{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{3} \right) \right)=-\dfrac{\pi }{3}\]
Thus our answer to the problem is \[-\dfrac{\pi }{3}\] .

Note: In these types of questions we need to be very careful about the domain and range of the given trigonometric function. The angle provided must be always less than \[\dfrac{\pi }{2}\] , if not, we need to convert it to an angle less than \[\dfrac{\pi }{2}\] , by making it a summation of an integral multiple of \[\pi \] and the remaining angle. The problem can also be done graphically by plotting it in a graph paper and analysing it.