Exercise 2.1 is based on the basic concepts of inverse trigonometry. The domain and range of the inverse trigonometric function (principle value branches) is the main concept of the chapter. It is recommended that you revise them before starting the exercise from this page. Also the basic definition of inverse trigonometric function should be known before starting this exercise.
This exercise forms the basics of inverse trigonometry and also is used in solving problems of Calculus.
Keeping this in mind we have given solution of Exercise 2.1 in full details so that students can have strong basics which will help them solve questions of further exercises.
Q1: Find the principal value of \sin^{-1}\left(-\frac{1}{2}\right).
Solution:
Step 1: Set up the equation:
Let’s start by assigning a variable to our function, just like we do in our standard textbook examples.
y = \sin^{-1}\left(-\frac{1}{2}\right)
Convert to a standard trigonometric equation:
By the definition of inverse functions, this means:
\sin y = -\frac{1}{2}
Step 2: Identify the corresponding angle:
We know from our standard trigonometric tables that
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}.
Because the sine function is negative in the fourth quadrant, we can write:
\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}
Step 3: Check the principal value branch:
We must always check our answer against the restricted range.
The range of the principal value branch for \sin^{-1} is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
Step 4: Final Conclusion:
Since -\frac{\pi}{6} falls perfectly within the interval \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], we have our answer.
The principal value of \sin^{-1}\left(-\frac{1}{2}\right) is -\frac{\pi}{6}.
Final Answer: \boxed{-\frac{\pi}{6}}.
Q2: Find the principal value of \cos^{-1}\left(\frac{\sqrt{3}}{2}\right).
Solution:
Step 1: Set up the equation:
Again, we will assign a variable to our expression.
y = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)
Convert to a standard trigonometric equation:
\cos y = \frac{\sqrt{3}}{2}
Step 2: Identify the corresponding angle:
We know from our basic trigonometry that:
\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}
Step 3: Check the principal value branch:
The range of the principal value branch for the \cos^{-1} function is [0, \pi].
Step 4: Final Conclusion:
Since \frac{\pi}{6} lies within the interval [0, \pi], it satisfies our condition.
The principal value of \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) is \frac{\pi}{6}.
Final Answer: \boxed{ \frac{\pi}{6}}
Q3: Find the principal value of \operatorname{cosec}^{-1}(2).
Solution:
Step 1: Set up the equation:
Let’s start by assigning a variable to our function, just like we do in our standard textbook examples.
y = \operatorname{cosec}^{-1}(2)
Convert to a standard trigonometric equation:
By the definition of inverse functions, this means:
\operatorname{cosec} y = 2
Step 2: Identify the corresponding angle:
We know from our standard trigonometric tables that \operatorname{cosec}\left(\frac{\pi}{6}\right) = 2.
Step 3: Check the principal value branch:
We must always check our answer against the restricted range.
The principal value branch of \operatorname{cosec}^{-1} is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}.
Step 4: Final Conclusion:
Since \frac{\pi}{6} falls perfectly within the interval \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}, we have our answer.
The principal value of \operatorname{cosec}^{-1}(2) is \frac{\pi}{6}.
Final Answer: \boxed{\frac{\pi}{6}}
Q4: Find the principal value of \tan^{-1}(-\sqrt{3}).
Solution:
Step 1: Set up the equation:
y = \tan^{-1}(-\sqrt{3})
Convert to a standard trigonometric equation:
\tan y = -\sqrt{3}
Step 2: Identify the corresponding angle:
We know that \tan\left(\frac{\pi}{3}\right) = \sqrt{3}.
Because the tangent function is negative in the fourth quadrant, we can write:
\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}
Step 3: Check the principal value branch:
The range of the principal value branch for \tan^{-1} is \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).
Step 4: Final Conclusion:
Since -\frac{\pi}{3} falls perfectly within the interval \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), we have our answer.
The principal value of \tan^{-1}(-\sqrt{3}) is -\frac{\pi}{3}.
Final Answer: \boxed{-\frac{\pi}{3}}
Q5: Find the principal value of \cos^{-1}\left(-\frac{1}{2}\right).
Solution:
Step 1: Set up the equation:
y = \cos^{-1}\left(-\frac{1}{2}\right)
Convert to a standard trigonometric equation:
\cos y = -\frac{1}{2}
Step 2: Identify the corresponding angle:
We know that \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}.
Because the cosine function is negative in the second quadrant, we subtract our angle from \pi:
\cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}
Step 3: Check the principal value branch:
The range of the principal value branch for \cos^{-1} is [0, \pi].
Step 4: Final Conclusion:
Since \frac{2\pi}{3} falls perfectly within the interval [0, \pi], we have our answer.
The principal value of \cos^{-1}\left(-\frac{1}{2}\right) is \frac{2\pi}{3}.
Final Answer: \boxed{\frac{2\pi}{3}}
Q6: Find the principal value of \tan^{-1}(-1).
Solution:
Step 1: Set up the equation:
y = \tan^{-1}(-1)
Convert to a standard trigonometric equation:
\tan y = -1
Step 2: Identify the corresponding angle:
We know that \tan\left(\frac{\pi}{4}\right) = 1.
Moving to the negative angle for tangent:
\tan\left(-\frac{\pi}{4}\right) = -1
Step 3: Check the principal value branch:
The range of the principal value branch for \tan^{-1} is \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).
Step 4: Final Conclusion:
Since -\frac{\pi}{4} lies within the interval \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), we have our answer.
The principal value of \tan^{-1}(-1) is -\frac{\pi}{4}.
Final Answer: \boxed{-\frac{\pi}{4}}
Q7: Find the principal value of \sec^{-1}\left(\frac{2}{\sqrt{3}}\right).
Solution:
Step 1: Set up the equation:
y = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)
Convert to a standard trigonometric equation:
\sec y = \frac{2}{\sqrt{3}}
Step 2: Identify the corresponding angle:
We know from our trigonometric tables that \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}.
Step 3: Check the principal value branch:
The range of the principal value branch for \sec^{-1} is [/latex][0, \pi] – \left\{\frac{\pi}{2}\right\}[/latex].
Step 4: Final Conclusion:
Since \frac{\pi}{6} falls perfectly within the restricted interval, we have our answer.
The principal value of \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) is \frac{\pi}{6}.
Final Answer: \boxed{\frac{\pi}{6}}
Q8: Find the principal value of \cot^{-1}(\sqrt{3}).
Solution:
Step 1: Set up the equation:
y = \cot^{-1}(\sqrt{3})
Convert to a standard trigonometric equation:
\cot y = \sqrt{3}
Step 2: Identify the corresponding angle:
We know that \cot\left(\frac{\pi}{6}\right) = \sqrt{3}.
Step 3: Check the principal value branch:
The range of the principal value branch for \cot^{-1} is (0, \pi).
Step 4: Final Conclusion:
Since \frac{\pi}{6} is within the interval (0, \pi), we have our answer.
The principal value of \cot^{-1}(\sqrt{3}) is \frac{\pi}{6}.
Final Answer: \boxed{\frac{\pi}{6}}
Q9: Find the principal value of \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right).
Solution:
Step 1: Set up the equation:
y = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)
Convert to a standard trigonometric equation:
\cos y = -\frac{1}{\sqrt{2}}
Step 2: Identify the corresponding angle:
We know \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}.
Finding the negative value in the second quadrant:
\cos\left(\pi - \frac{\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}
Step 3: Check the principal value branch:
The range of the principal value branch for \cos^{-1} is [0, \pi].
Step 4: Final Conclusion:
Since \frac{3\pi}{4} falls perfectly within [0, \pi], we have our answer.
The principal value of \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) is \frac{3\pi}{4}.
Final Answer: \boxed{\frac{3\pi}{4}}
Q10: Find the principal value of \operatorname{cosec}^{-1}(-\sqrt{2}).
Solution:
Step 1: Set up the equation:
y = \operatorname{cosec}^{-1}(-\sqrt{2})
Convert to a standard trigonometric equation:
\operatorname{cosec} y = -\sqrt{2}
Step 2: Identify the corresponding angle:
We know \operatorname{cosec}\left(\frac{\pi}{4}\right) = \sqrt{2}.
Moving to the negative angle:
\operatorname{cosec}\left(-\frac{\pi}{4}\right) = -\sqrt{2}
Step 3: Check the principal value branch:
The principal value branch of \operatorname{cosec}^{-1} is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}.
Step 4: Final Conclusion:
Since -\frac{\pi}{4} falls within the restricted range, we have our answer.
The principal value of \operatorname{cosec}^{-1}(-\sqrt{2}) is -\frac{\pi}{4}.
Final Answer: \boxed{-\frac{\pi}{4}}
Q11: Find the value of \tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right).
Solution:
Step 1: Identify the principal values of each individual term:
Using our standard table of principal value branches:
\tan^{-1}(1) = \frac{\pi}{4}
\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}
\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}
Step 2: Substitute these values back into the original expression:
\frac{\pi}{4} + \frac{2\pi}{3} + \left(-\frac{\pi}{6}\right)
Step 3: Calculate the final sum:
Find a common denominator of 12 for the fractions:
\frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12}
Simplifying the fraction gives us \frac{3\pi}{4}.
Final Answer: \boxed{\frac{3\pi}{4}}
Q12: Find the value of \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right).
Solution:
Step 1: Identify the principal values of each individual term:
Keeping the principal branches in mind:
\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}
\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}
Step 2: Substitute these values back into the original expression:
\frac{\pi}{3} + 2\left(\frac{\pi}{6}\right)
Step 3: Calculate the final sum:
Multiply and add the terms:
\frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}
Final Answer: \boxed{\frac{2\pi}{3}}
Q13: If \sin^{-1}x = y, then
(A) 0 \le y \le \pi
(B) -\frac{\pi}{2} \le y \le \frac{\pi}{2}
(C) 0 < y < \pi
(D) -\frac{\pi}{2} < y < \frac{\pi}{2}
Solution:
Step 1: Understand the definition:
The question is asking us to identify the domain or range of the given function.
When we write y = \sin^{-1}x, y represents the range (or principal value branch) of the inverse sine function.
Step 2: Recall the principal value branch:
If we refer back to the standard table of inverse trigonometric functions, the principal value branch for y = \sin^{-1}x is the closed interval \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
Step 3: Final Conclusion:
Because the interval is closed, it includes the endpoints.
Therefore, the mathematically correct inequality is -\frac{\pi}{2} \le y \le \frac{\pi}{2}. This perfectly matches option (B).
Final Answer: \boxed{\text{Option (B)}}
Q14: \tan^{-1}\sqrt{3} - \sec^{-1}(-2) is equal to
(A) \pi
(B) -\frac{\pi}{3}
(C) \frac{\pi}{3}
(D) \frac{2\pi}{3}
Solution:
Step 1: Find the principal value of the first term:
Let’s evaluate \tan^{-1}\sqrt{3}.
We know from our trigonometric tables that \tan\left(\frac{\pi}{3}\right) = \sqrt{3}.
Since \frac{\pi}{3} falls comfortably within the principal branch of \tan^{-1}, which is \left(-\frac{\pi}{2}, \frac{\pi}{2}\right),
we have:
\tan^{-1}\sqrt{3} = \frac{\pi}{3}
Step 2: Find the principal value of the second term:
Now, let y = \sec^{-1}(-2), which means \sec y = -2.
We know \sec\left(\frac{\pi}{3}\right) = 2.
Because the secant function is negative in the second quadrant, we subtract the angle from \pi: \pi - \frac{\pi}{3} = \frac{2\pi}{3}.
The principal branch for \sec^{-1} is [0, \pi] - \left\{\frac{\pi}{2}\right\}, and \frac{2\pi}{3} is perfectly valid within this range.
So: \sec^{-1}(-2) = \frac{2\pi}{3}
Step 3: Calculate the final expression:
Now, we just substitute our two principal values back into the original expression given in the question:
\frac{\pi}{3} - \frac{2\pi}{3}
= -\frac{\pi}{3}
Step 4: Final Conclusion:
The calculated result is -\frac{\pi}{3}, which corresponds to option (B).
Final Answer: \boxed{\text{Option (B)}}