Question

How do you evaluate ${{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)?$

Answer 1

Hint: We will deal with the inner bracket first. We will use an important trigonometric identity to find the value of the given trigonometric function. That is given by, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.$ Then we will consider the whole function.

Complete step by step solution:
Let us take the given trigonometric function ${{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)$ into consideration.
Take the value inside the inner bracket, $\dfrac{11\pi }{10}.$
We can write this as,
$\Rightarrow \dfrac{11\pi }{10}=\dfrac{\left( 10+1 \right)\pi }{10}.$
This can be written as,
$\Rightarrow \dfrac{11\pi }{10}=\dfrac{10\pi +\pi }{10}.$
From this we will get,
$\Rightarrow \dfrac{11\pi }{10}=\dfrac{10\pi }{10}+\dfrac{\pi }{10}.$
Cancelling $10$ from the first fraction in the left-hand side,
$\Rightarrow \dfrac{11\pi }{10}=\pi +\dfrac{\pi }{10}.$
Now we are supposed to consider the term inside the outer bracket, $\sin \dfrac{11\pi }{10}.$
Using the above obtained equation to write,
$\Rightarrow \sin \dfrac{11\pi }{10}=\sin \left( \pi +\dfrac{\pi }{10} \right).$
Hence, we can see that the right-hand side of the above equation is in the form $\sin \left( A+B \right).$
So, let us recall the trigonometric formula, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B.$
Using this formula will help us to find the value of the given trigonometric function.
Here, $A=\pi $ and $B=\dfrac{\pi }{10}.$
We will get the following equation,
$\sin \dfrac{11\pi }{10}=\sin \left( \pi +\dfrac{\pi }{10} \right)=\sin \pi \cos \dfrac{\pi }{10}+\cos \pi \sin \dfrac{\pi }{10}.$
We know the values of the functions in the above obtained equation.
It is known that, $\sin \pi =0$ and $\cos \pi =-1.$
Therefore, the first summand in the right-hand side becomes,
$\Rightarrow \sin \pi \cos \dfrac{\pi }{10}=0\times \cos \dfrac{\pi }{10}=0.$
The second summand in the right-hand side becomes,
$\Rightarrow \cos \pi \sin \dfrac{\pi }{10}=\left( -1 \right)\sin \dfrac{\pi }{10}=-\sin \dfrac{\pi }{10}.$
And hence, the equation will become as,
$\Rightarrow \sin \dfrac{11\pi }{10}=0\times \cos \dfrac{\pi }{10}+\left( -1 \right)\sin \dfrac{\pi }{10}.$
That is,
$\Rightarrow \sin \dfrac{11\pi }{10}=0+\left( -1 \right)\sin \dfrac{\pi }{10}=-\sin \dfrac{\pi }{10}.$
Now we consider the whole function we are given with, then
$\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)={{\sin }^{-1}}\left( -\sin \left( \dfrac{\pi }{10} \right) \right).$
Since the sine function is odd, $\sin \left( -\theta \right)=-\sin \theta $ and ${{\sin }^{-1}}\left( -\theta \right)=-{{\sin }^{-1}}\theta $
We will get,
$\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)=-{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{10} \right) \right).$
Now we will have,
$\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)=-\left( \dfrac{\pi }{10} \right).$ Since, ${{\sin }^{-1}}\left( \sin \theta \right)=\theta ,\,\,\,\,\,\,\,\,\,\,\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ and also, $\dfrac{\pi }{10}\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].$
Hence, ${{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)=-\left( \dfrac{\pi }{10} \right).$

Note: Note that there is another method:
Consider the trigonometric function we are given with,
$\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)$
Now,
\[\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)={{\sin }^{-1}}\left( \sin \left( \dfrac{10\pi +\pi }{10} \right) \right).\]
We repeat the procedure inside the inner bracket,
$\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)={{\sin }^{-1}}\left( \sin \left( \dfrac{10\pi }{10}+\dfrac{\pi }{10} \right) \right).$
Then, we get
$\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)={{\sin }^{-1}}\left( \sin \left( \pi +\dfrac{\pi }{10} \right) \right).$
In the third quadrant, sine is negative.
That is, $\sin \left( \pi +\theta \right)=-\sin \theta .$
Now the above equation becomes,
$\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)={{\sin }^{-1}}\left( -\sin \left( \dfrac{\pi }{10} \right) \right).$
We use the identity ${{\sin }^{-1}}\left( -\theta \right)=-{{\sin }^{-1}}\theta $,
$\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)=-{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{10} \right) \right).$
Now we use another identity, ${{\sin }^{-1}}\left( \sin \theta \right)=\theta ,$ if $\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].$
\[\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)=-\dfrac{\pi }{10}.\]