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

How do you evaluate sin⁡−1(sin⁡(11π10))?{{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)?

Explanation

Solution

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⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B.\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(sin⁡(11π10)){{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right) into consideration.
Take the value inside the inner bracket, 11π10.\dfrac{11\pi }{10}.
We can write this as,
⇒11π10=(10+1)π10.\Rightarrow \dfrac{11\pi }{10}=\dfrac{\left( 10+1 \right)\pi }{10}.
This can be written as,
⇒11π10=10π+π10.\Rightarrow \dfrac{11\pi }{10}=\dfrac{10\pi +\pi }{10}.
From this we will get,
⇒11π10=10π10+π10.\Rightarrow \dfrac{11\pi }{10}=\dfrac{10\pi }{10}+\dfrac{\pi }{10}.
Cancelling 1010 from the first fraction in the left-hand side,
⇒11π10=π+π10.\Rightarrow \dfrac{11\pi }{10}=\pi +\dfrac{\pi }{10}.
Now we are supposed to consider the term inside the outer bracket, sin⁡11π10.\sin \dfrac{11\pi }{10}.
Using the above obtained equation to write,
⇒sin⁡11π10=sin⁡(π+π10).\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⁡(A+B).\sin \left( A+B \right).
So, let us recall the trigonometric formula, sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B.\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=πA=\pi and B=π10.B=\dfrac{\pi }{10}.
We will get the following equation,
sin⁡11π10=sin⁡(π+π10)=sin⁡πcos⁡π10+cos⁡πsin⁡π10.\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⁡π=0\sin \pi =0 and cos⁡π=−1.\cos \pi =-1.
Therefore, the first summand in the right-hand side becomes,
⇒sin⁡πcos⁡π10=0×cos⁡π10=0.\Rightarrow \sin \pi \cos \dfrac{\pi }{10}=0\times \cos \dfrac{\pi }{10}=0.
The second summand in the right-hand side becomes,
⇒cos⁡πsin⁡π10=(−1)sin⁡π10=−sin⁡π10.\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,
⇒sin⁡11π10=0×cos⁡π10+(−1)sin⁡π10.\Rightarrow \sin \dfrac{11\pi }{10}=0\times \cos \dfrac{\pi }{10}+\left( -1 \right)\sin \dfrac{\pi }{10}.
That is,
⇒sin⁡11π10=0+(−1)sin⁡π10=−sin⁡π10.\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
⇒sin⁡−1(sin⁡(11π10))=sin⁡−1(−sin⁡(π10)).\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⁡(−θ)=−sin⁡θ\sin \left( -\theta \right)=-\sin \theta and sin⁡−1(−θ)=−sin⁡−1θ{{\sin }^{-1}}\left( -\theta \right)=-{{\sin }^{-1}}\theta
We will get,
⇒sin⁡−1(sin⁡(11π10))=−sin⁡−1(sin⁡(π10)).\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,
⇒sin⁡−1(sin⁡(11π10))=−(π10).\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)=-\left( \dfrac{\pi }{10} \right). Since, sin⁡−1(sin⁡θ)=θ,          θ∈[−π2,π2]{{\sin }^{-1}}\left( \sin \theta \right)=\theta ,\,\,\,\,\,\,\,\,\,\,\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] and also, π10∈[−π2,π2].\dfrac{\pi }{10}\in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
Hence, sin⁡−1(sin⁡(11π10))=−(π10).{{\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,
⇒sin⁡−1(sin⁡(11π10))\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)
Now,
⇒sin⁡−1(sin⁡(11π10))=sin⁡−1(sin⁡(10π+π10)).\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,
⇒sin⁡−1(sin⁡(11π10))=sin⁡−1(sin⁡(10π10+π10)).\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
⇒sin⁡−1(sin⁡(11π10))=sin⁡−1(sin⁡(π+π10)).\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⁡(π+θ)=−sin⁡θ.\sin \left( \pi +\theta \right)=-\sin \theta .
Now the above equation becomes,
⇒sin⁡−1(sin⁡(11π10))=sin⁡−1(−sin⁡(π10)).\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(−θ)=−sin⁡−1θ{{\sin }^{-1}}\left( -\theta \right)=-{{\sin }^{-1}}\theta ,
⇒sin⁡−1(sin⁡(11π10))=−sin⁡−1(sin⁡(π10)).\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(sin⁡θ)=θ,{{\sin }^{-1}}\left( \sin \theta \right)=\theta , if θ∈[−π2,π2].\theta \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
⇒sin⁡−1(sin⁡(11π10))=−π10.\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{11\pi }{10} \right) \right)=-\dfrac{\pi }{10}.