Answer
418.5k+ views
Hint: We will first find the principal value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$. Then we will find the principal value of ${{\cos }^{-1}}\left( 0 \right)$. We will add these two values to get the required answer. The principal value of an inverse trigonometric function at a point $x$ is the value of the inverse function at the point $x$, which lies in the range of the principal branch.
Complete step by step answer:
We know that the principle interval of ${{\sin }^{-1}}\theta $ is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. Let ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=x$.
Therefore, we now know that $\sin x=\dfrac{1}{2}$. This implies that $x=\dfrac{\pi }{6}$. Since $\dfrac{\pi }{6}\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, we can say that $x=\dfrac{\pi }{6}$ is the principle value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$.
Now, we will find the principal value of ${{\cos }^{-1}}\left( 0 \right)$. We know that the principal interval of ${{\cos }^{-1}}\theta $ is $\left[ 0,\pi \right]$. Let ${{\cos }^{-1}}\left( 0 \right)=y$. Therefore, we have $\cos y=0$. So, we get $y=\dfrac{\pi }{2}$. Since $\dfrac{\pi }{2}\in \left[ 0,\pi \right]$, we conclude that $y=\dfrac{\pi }{2}$ is the principle value of ${{\cos }^{-1}}\left( 0 \right)$.
Now, we can substitute the values of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ and ${{\cos }^{-1}}\left( 0 \right)$ to find the required answer in the following manner,
${{\sin }^{-1}}\left( \dfrac{1}{2} \right)+{{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{6}+\dfrac{\pi }{2}=\dfrac{\pi +3\pi }{6}=\dfrac{4\pi }{6}=\dfrac{2\pi }{3}$.
Therefore, the answer is $\dfrac{2\pi }{3}$.
Note: There are multiple angles which have the same values for the trigonometric functions. The concept of principle value is essential so that values can be standardized for solving questions. Apart from trigonometry, this concept also arises in certain functions that involve complex numbers. Similar to trigonometric functions, there are identities and relations involving inverse trigonometric functions. We should be familiar with these identities as it will be useful for simplifying equations involving inverse trigonometric functions.
Complete step by step answer:
We know that the principle interval of ${{\sin }^{-1}}\theta $ is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. Let ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=x$.
Therefore, we now know that $\sin x=\dfrac{1}{2}$. This implies that $x=\dfrac{\pi }{6}$. Since $\dfrac{\pi }{6}\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, we can say that $x=\dfrac{\pi }{6}$ is the principle value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$.
Now, we will find the principal value of ${{\cos }^{-1}}\left( 0 \right)$. We know that the principal interval of ${{\cos }^{-1}}\theta $ is $\left[ 0,\pi \right]$. Let ${{\cos }^{-1}}\left( 0 \right)=y$. Therefore, we have $\cos y=0$. So, we get $y=\dfrac{\pi }{2}$. Since $\dfrac{\pi }{2}\in \left[ 0,\pi \right]$, we conclude that $y=\dfrac{\pi }{2}$ is the principle value of ${{\cos }^{-1}}\left( 0 \right)$.
Now, we can substitute the values of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ and ${{\cos }^{-1}}\left( 0 \right)$ to find the required answer in the following manner,
${{\sin }^{-1}}\left( \dfrac{1}{2} \right)+{{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{6}+\dfrac{\pi }{2}=\dfrac{\pi +3\pi }{6}=\dfrac{4\pi }{6}=\dfrac{2\pi }{3}$.
Therefore, the answer is $\dfrac{2\pi }{3}$.
Note: There are multiple angles which have the same values for the trigonometric functions. The concept of principle value is essential so that values can be standardized for solving questions. Apart from trigonometry, this concept also arises in certain functions that involve complex numbers. Similar to trigonometric functions, there are identities and relations involving inverse trigonometric functions. We should be familiar with these identities as it will be useful for simplifying equations involving inverse trigonometric functions.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)