Chapter 2 Inverse Trigonometric Functions

March 10, 2024
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Chapter 2 Inverse Trigonometric Functions

EXERCISE 2.1

Question 1:

Find the principal value of $\displaystyle {{\operatorname{Sin}}^{{-1}}}\left( {\frac{1}{2}} \right)$

Solution: Let, $\displaystyle {{\operatorname{Sin}}^{{-1}}}\left( {\frac{1}{2}} \right)=y$ Hence, $\displaystyle \sin y=\left( {-\frac{1}{2}} \right)$ = $\displaystyle -\sin \left( {\frac{\pi }{6}} \right)$ = $\displaystyle \sin \left( {-\frac{\pi }{6}} \right)$ Range of the principal value of $\displaystyle {{\sin }^{{-1}}}\left( x \right)$ is $\displaystyle \left( {-\frac{\pi }{2},\frac{\pi }{2}} \right)$ Thus, principal value of $\displaystyle {{\sin }^{{-1}}}\left( {-\frac{1}{2}} \right)=\left( {-\frac{\pi }{6}} \right)$

Find the principal value of .

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Chapter 2 Inverse Trigonometric Functions

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