In this chapter, we will learn and solve the questions of NCERT exercise of inverse trigonometric functions 2.1.
We know that a function has an inverse only when it is bijective, i.e., both one-one and onto.
However, trigonometric functions are generally not bijective over their entire domains because they repeat their values periodically.
For example:
\sin 0=\sin \pi=\sin 2\pi=0
To make trigonometric functions invertible, we restrict their domains to certain intervals where they become bijective. These intervals are called principal value branches.
The inverses obtained on these restricted intervals are called inverse trigonometric functions.
Key Concepts
1. Inverse Function
A function has an inverse only if it is bijective, i.e., both one-one and onto.
If f:A\to B is bijective, then its inverse is denoted by f^{-1}:B\to A .
Example:
If f(x)=2x+3 , then
f^{-1}(x)=\frac{x-3}{2}
2. Trigonometric Functions are not Invertible in General
Trigonometric functions are generally not invertible over their entire domains because they are not one-one.
They repeat their values periodically.
Example:
\sin 0=\sin \pi=\sin 2\pi=0
Thus, different values of x give the same value of \sin x.
3. Principal Value Branches
To make trigonometric functions invertible, their domains are restricted to certain intervals where they become bijective.
These restricted intervals are called principal value branches.
The commonly used principal value branches are:
- \sin x:\ \left[-\frac{\pi}{2},\frac{\pi}{2}\right]
- \cos x:\ [0,\pi]
- \tan x:\ \left(-\frac{\pi}{2},\frac{\pi}{2}\right)
4. Inverse Trigonometric Functions
The inverses obtained after restricting the domains of trigonometric functions are called inverse trigonometric functions.
They are denoted by:
\sin^{-1}x,\ \cos^{-1}x,\ \tan^{-1}x,\ \cot^{-1}x,\ \sec^{-1}x,\ \csc^{-1}x
Example:
If y=\sin x , then
x=\sin^{-1}y
5. Domain and Range of Inverse Trigonometric Functions
For inverse functions, domain and range are interchanged.
Thus, the domain of an inverse trigonometric function is the range of the corresponding trigonometric function, while its range is the principal value branch.
Examples:
- \sin^{-1}x
Domain: [-1,1]
Range: \left[-\frac{\pi}{2},\frac{\pi}{2}\right]
- \cos^{-1}x
Domain: [-1,1]
Range: [0,\pi]
- \tan^{-1}x
Domain: \mathbb{R}
Range: \left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Domains, Ranges and Principal Value Branches
The following table gives a quick comparison of the domains and ranges of trigonometric functions along with the principal value branches of their inverse functions. Learning these standard intervals carefully is very important for solving inverse trigonometric function questions correctly.
| T | Dom | R=D | PVB | Inv |
|---|---|---|---|---|
| \sin | \mathbb{R} | [-1,1] | [-\frac{\pi}{2},\frac{\pi}{2}] | \sin^{-1} |
| \cos | \mathbb{R} | [-1,1] | [0,\pi] | \cos^{-1} |
| \tan | \mathbb{R}\setminus\{\frac{(2n+1)\pi}{2}\} | \mathbb{R} | (-\frac{\pi}{2},\frac{\pi}{2}) | \tan^{-1} |
| \cot | \mathbb{R}\setminus\{n\pi\} | \mathbb{R} | (0,\pi) | \cot^{-1} |
| \sec | \mathbb{R}\setminus\{\frac{(2n+1)\pi}{2}\} | \mathbb{R}\setminus(-1,1) | [0,\pi]\setminus\{\frac{\pi}{2}\} | \sec^{-1} |
| \csc | \mathbb{R}\setminus\{n\pi\} | \mathbb{R}\setminus(-1,1) | [-\frac{\pi}{2},\frac{\pi}{2}]\setminus\{0\} | \csc^{-1} |
Here, n\in\mathbb{Z} and PVB denotes Principal Value Branch.
Let us now solve all the NCERT questions step by step in this exercise of Inverse Trigonometric Functions 2.1.
Question 1: Inverse Trigonometric Functions 2.1
1. Find the principal value of \sin^{-1}\left(-\frac{1}{2}\right)
Solution
Let
\sin^{-1}\left(-\frac{1}{2}\right)=y
Then,
\sin y=-\frac{1}{2}
Now, sine is negative in the 3rd and 4th quadrants. Also,
\sin\frac{\pi}{6}=\frac{1}{2}
Therefore, angles whose sine is -\frac{1}{2} can be written as
\sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2}
or
\sin\left(\pi+\frac{\pi}{6}\right)=-\frac{1}{2}
But \sin^{-1}x gives only that value which belongs to its principal value branch
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Now, -\frac{\pi}{6} lies in this interval, whereas \frac{7\pi}{6} does not.
Therefore,
y=-\frac{\pi}{6}
Hence,
\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}
Question 2: Principal Value
2. Find the principal value of \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)
Solution
Let
\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=y
Then,
\cos y=\frac{\sqrt{3}}{2}
Now, cosine is positive in the 1st and 4th quadrants. Also,
\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}
Therefore, angles whose cosine is \frac{\sqrt{3}}{2} can be written as
\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}
or
\cos\left(2\pi-\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}
But \cos^{-1}x gives only that value which belongs to its principal value branch
[0,\pi]
Now, \frac{\pi}{6} lies in this interval, whereas \frac{11\pi}{6} does not.
Therefore,
y=\frac{\pi}{6}
Hence,
\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}
You may already be following Maths Better for NCERT Solutions for the topics like Matrices, Determinants, and Relations and Functions. Similarly, this exercise on Inverse Trigonometric Functions for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills.
Now, let’s move on to the next question of Inverse Trigonometric Functions 2.1.
Question 3: Inverse Trigonometric Functions 2.1
3. Find the principal value of \cosec^{-1}(2)
Solution
To find \cosec^{-1}(2), we first convert it into a sine inverse function, since
\cosec^{-1}x=\sin^{-1}\left(\frac{1}{x}\right)
Therefore,
\cosec^{-1}(2)=\sin^{-1}\left(\frac{1}{2}\right)
Let
\sin^{-1}\left(\frac{1}{2}\right)=y
Then,
\sin y=\frac{1}{2}
Now, sine is positive in the 1st and 2nd quadrants. Also,
\sin\frac{\pi}{6}=\frac{1}{2}
Therefore, angles whose sine is \frac{1}{2} can be written as
\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}
or
\sin\left(\pi-\frac{\pi}{6}\right)=\frac{1}{2}
But \sin^{-1}x gives only that value which belongs to its principal value branch
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Now, \frac{\pi}{6} lies in this interval, whereas \frac{5\pi}{6} does not.
Therefore,
y=\frac{\pi}{6}
Hence,
\cosec^{-1}(2)=\frac{\pi}{6}
Alternative Method:
We can also solve it directly using cosecant.
Let
\cosec^{-1}(2)=y
Then,
\cosec y=2
Now,
\cosec\frac{\pi}{6}=2
and \frac{\pi}{6} lies in the principal value branch of \cosec^{-1}x.
Therefore,
\cosec^{-1}(2)=\frac{\pi}{6}
Question 4: Exercise 2.1
4. Find the principal value of \tan^{-1}(-\sqrt{3})
Solution
Let
\tan^{-1}(-\sqrt{3})=y
Then,
\tan y=-\sqrt{3}
Now, tangent is negative in the 2nd and 4th quadrants. Also,
\tan\frac{\pi}{3}=\sqrt{3}
Therefore, angles whose tangent is -\sqrt{3} can be written as
\tan\left(-\frac{\pi}{3}\right)=-\sqrt{3}
or
\tan\left(\pi-\frac{\pi}{3}\right)=-\sqrt{3}
But \tan^{-1}x gives only that value which belongs to its principal value branch
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Now, -\frac{\pi}{3} lies in this interval, whereas \frac{2\pi}{3} does not.
Therefore,
y=-\frac{\pi}{3}
Hence,
\tan^{-1}(-\sqrt{3})=-\frac{\pi}{3}
Question 5: Inverse Trigonometric Functions Ex. 2.1
5. Find the principal value of \cos^{-1}\left(-\frac{1}{2}\right)
Solution
Let
\cos^{-1}\left(-\frac{1}{2}\right)=y
Then,
\cos y=-\frac{1}{2}
Now, cosine is negative in the 2nd and 3rd quadrants. Also,
\cos\frac{\pi}{3}=\frac{1}{2}
Therefore, angles whose cosine is -\frac{1}{2} can be written as
\cos\left(\pi-\frac{\pi}{3}\right)=-\frac{1}{2}
or
\cos\left(\pi+\frac{\pi}{3}\right)=-\frac{1}{2}
But \cos^{-1}x gives only that value which belongs to its principal value branch
[0,\pi]
Now, \frac{2\pi}{3} lies in this interval, whereas \frac{4\pi}{3} does not.
Therefore,
y=\frac{2\pi}{3}
Hence,
\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}
While solving inverse trigonometric function questions, always remember that many angles may satisfy a trigonometric equation, but the final answer must belong to the principal value branch of the given inverse trigonometric function.
Question 6: Inverse Trigonometric Functions 2.1
6. Find the principal value of \tan^{-1}(-1)
Solution
Let
\tan^{-1}(-1)=y
Then,
\tan y=-1
Now, tangent is negative in the 2nd and 4th quadrants. Also,
\tan\frac{\pi}{4}=1
Therefore, angles whose tangent is -1 can be written as
\tan\left(-\frac{\pi}{4}\right)=-1
or
\tan\left(\pi-\frac{\pi}{4}\right)=-1
But \tan^{-1}x gives only that value which belongs to its principal value branch
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Now, -\frac{\pi}{4} lies in this interval, whereas \frac{3\pi}{4} does not.
Therefore,
y=-\frac{\pi}{4}
Hence,
\tan^{-1}(-1)=-\frac{\pi}{4}
Question 7: Ex 2.1 Inverse Trigonometric Functions
7. Find the principal value of \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)
Solution
To find \sec^{-1}\left(\frac{2}{\sqrt{3}}\right), we first convert it into a cosine inverse function, since
\sec^{-1}x=\cos^{-1}\left(\frac{1}{x}\right)
Therefore,
\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)=\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)
Let
\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=y
Then,
\cos y=\frac{\sqrt{3}}{2}
Now, cosine is positive in the 1st and 4th quadrants. Also,
\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}
Therefore, angles whose cosine is \frac{\sqrt{3}}{2} can be written as
\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}
or
\cos\left(2\pi-\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}
But \cos^{-1}x gives only that value which belongs to its principal value branch
[0,\pi]
Now, \frac{\pi}{6} lies in this interval, whereas \frac{11\pi}{6} does not.
Therefore,
y=\frac{\pi}{6}
Hence,
\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter, to help you understand the concepts more visually.
Now, let’s move to the next question.
Question 8: Inverse Trigonometric Functions 2.1
8. Find the principal value of \cot^{-1}(\sqrt{3})
Solution
Let
\cot^{-1}(\sqrt{3})=y
Then,
\cot y=\sqrt{3}
Now, cotangent is positive in the 1st and 3rd quadrants. Also,
\cot\frac{\pi}{6}=\sqrt{3}
Therefore, angles whose cotangent is \sqrt{3} can be written as
\cot\left(\frac{\pi}{6}\right)=\sqrt{3}
or
\cot\left(\pi+\frac{\pi}{6}\right)=\sqrt{3}
But \cot^{-1}x gives only that value which belongs to its principal value branch
(0,\pi)
Now, \frac{\pi}{6} lies in this interval, whereas \frac{7\pi}{6} does not.
Therefore,
y=\frac{\pi}{6}
Hence,
\cot^{-1}(\sqrt{3})=\frac{\pi}{6}
Converting inverse trigonometric expressions into standard trigonometric equations like \sin y=x or \cos y=x often makes the correct angle and principal value easier to identify.
Question 9: Inverse Trigonometric Functions 2.1
9. Find the principal value of \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)
Solution
Let
\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)=y
Then,
\cos y=-\frac{1}{\sqrt{2}}
Now, cosine is negative in the 2nd and 3rd quadrants. Also,
\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}
Therefore, angles whose cosine is -\frac{1}{\sqrt{2}} can be written as
\cos\left(\pi-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}
or
\cos\left(\pi+\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}
But \cos^{-1}x gives only that value which belongs to its principal value branch
[0,\pi]
Now, \frac{3\pi}{4} lies in this interval, whereas \frac{5\pi}{4} does not.
Therefore,
y=\frac{3\pi}{4}
Hence,
\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi}{4}
Question 10: Inverse Trigonometric Functions 2.1 Exercise
10. Find the principal value of \cosec^{-1}(-\sqrt{2})
Solution
To find \cosec^{-1}(-\sqrt{2}), we first convert it into a sine inverse function, since
\cosec^{-1}x=\sin^{-1}\left(\frac{1}{x}\right)
Therefore,
\cosec^{-1}(-\sqrt{2})=\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)
Let
\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)=y
Then,
\sin y=-\frac{1}{\sqrt{2}}
Now, sine is negative in the 3rd and 4th quadrants. Also,
\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}
Therefore, angles whose sine is -\frac{1}{\sqrt{2}} can be written as
\sin\left(-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}
or
\sin\left(\pi+\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}
But \sin^{-1}x gives only that value which belongs to its principal value branch
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Now, -\frac{\pi}{4} lies in this interval, whereas \frac{5\pi}{4} does not.
Therefore,
y=-\frac{\pi}{4}
Hence,
\cosec^{-1}(-\sqrt{2})=-\frac{\pi}{4}
There can be many angles having the same trigonometric value. Practising how to identify the correct angle using quadrant signs and principal value branches greatly improves conceptual understanding of inverse trigonometric functions.
Here’s the next question.
Question 11: Inverse Trigo Functions Ex. 2.1
11. Find the value of \tan^{-1}(1)+\cos^{-1}\left(-\frac{1}{2}\right)+\sin^{-1}\left(-\frac{1}{2}\right)
Solution
We first find the principal values of each inverse trigonometric function separately.
1. Finding \tan^{-1}(1)
Let
\tan^{-1}(1)=a
Then,
\tan a=1
Now, tangent is positive in the 1st and 3rd quadrants. Also,
\tan\frac{\pi}{4}=1
But \tan^{-1}x gives only that value which belongs to its principal value branch
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Therefore,
\tan^{-1}(1)=\frac{\pi}{4}
2. Finding \cos^{-1}\left(-\frac{1}{2}\right)
Let
\cos^{-1}\left(-\frac{1}{2}\right)=b
Then,
\cos b=-\frac{1}{2}
Now, cosine is negative in the 2nd and 3rd quadrants. Also,
\cos\frac{\pi}{3}=\frac{1}{2}
Therefore, the required principal value is
\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}
3. Finding \sin^{-1}\left(-\frac{1}{2}\right)
Let
\sin^{-1}\left(-\frac{1}{2}\right)=c
Then,
\sin c=-\frac{1}{2}
Now, sine is negative in the 3rd and 4th quadrants. Also,
\sin\frac{\pi}{6}=\frac{1}{2}
Therefore, the required principal value is
\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}
Substituting these values, we get
=\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}
=\frac{3\pi+8\pi-2\pi}{12}
=\frac{9\pi}{12}
=\frac{3\pi}{4}
Hence,
\tan^{-1}(1)+\cos^{-1}\left(-\frac{1}{2}\right)+\sin^{-1}\left(-\frac{1}{2}\right)=\frac{3\pi}{4}
Practice is the key to mastering Inverse Trigonometric Functions 2.1. Solve questions step by step and try to understand the logic behind the principal value. If you have any doubt, feel free to leave a comment.
Question 12: Inverse Trigonometric Functions 2.1
12. Find the value of \cos^{-1}\left(\frac{1}{2}\right)+2\sin^{-1}\left(\frac{1}{2}\right)
Solution
We first find the principal values of the inverse trigonometric functions separately.
1. Finding \cos^{-1}\left(\frac{1}{2}\right)
Let
\cos^{-1}\left(\frac{1}{2}\right)=a
Then,
\cos a=\frac{1}{2}
Now, cosine is positive in the 1st and 4th quadrants. Also,
\cos\frac{\pi}{3}=\frac{1}{2}
But \cos^{-1}x gives only that value which belongs to its principal value branch
[0,\pi]
Therefore,
\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}
2. Finding \sin^{-1}\left(\frac{1}{2}\right)
Let
\sin^{-1}\left(\frac{1}{2}\right)=b
Then,
\sin b=\frac{1}{2}
Now, sine is positive in the 1st and 2nd quadrants. Also,
\sin\frac{\pi}{6}=\frac{1}{2}
But \sin^{-1}x gives only that value which belongs to its principal value branch
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Therefore,
\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}
Substituting these values, we get
=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right)
=\frac{\pi}{3}+\frac{\pi}{3}
=\frac{2\pi}{3}
Hence,
\cos^{-1}\left(\frac{1}{2}\right)+2\sin^{-1}\left(\frac{1}{2}\right)=\frac{2\pi}{3}
Now, let’s move on to the next two MCQs of Inverse Trigonometric Functions 2.1.
Question 13: Inverse Trigonometric Functions – MCQ
13. 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
We are given
\sin^{-1}x=y
Now, whenever we see an inverse trigonometric function, the first thing to think about is its principal value branch, because inverse trigonometric functions return values only from their restricted interval.
For \sin^{-1}x, the principal value branch is
\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
Since y=\sin^{-1}x, the value of y must belong to this interval.
Therefore,
-\frac{\pi}{2}\le y\le \frac{\pi}{2}
Hence, the correct option is (B).
Question 14: Inverse Trigonometric Functions 2.1 – MCQ
14. \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
We first find the principal values of the inverse trigonometric functions separately.
1. Finding \tan^{-1}(\sqrt{3})
Let
\tan^{-1}(\sqrt{3})=a
Then,
\tan a=\sqrt{3}
Now, tangent is positive in the 1st and 3rd quadrants. Also,
\tan\frac{\pi}{3}=\sqrt{3}
But \tan^{-1}x gives only that value which belongs to its principal value branch
\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
Therefore,
\tan^{-1}(\sqrt{3})=\frac{\pi}{3}
2. Finding \sec^{-1}(-2)
We first convert it into a cosine inverse function:
\sec^{-1}x=\cos^{-1}\left(\frac{1}{x}\right)
Therefore,
\sec^{-1}(-2)=\cos^{-1}\left(-\frac{1}{2}\right)
Let
\cos^{-1}\left(-\frac{1}{2}\right)=b
Then,
\cos b=-\frac{1}{2}
Now, cosine is negative in the 2nd and 3rd quadrants. Also,
\cos\frac{\pi}{3}=\frac{1}{2}
Therefore, the required principal value is
\sec^{-1}(-2)=\frac{2\pi}{3}
Substituting these values, we get
=\frac{\pi}{3}-\frac{2\pi}{3}
=-\frac{\pi}{3}
Hence,
\tan^{-1}(\sqrt{3})-\sec^{-1}(-2)=-\frac{\pi}{3}
Therefore, the correct option is (B).
Common Mistakes to Avoid
- Confusing inverse with reciprocal: Students often think \sin^{-1}x means \frac{1}{\sin x}. This is incorrect. \sin^{-1}x means the inverse sine function, whereas \frac{1}{\sin x}=\cosec x.
- Ignoring principal value branches: Inverse trigonometric functions return only one value from their restricted interval. Always check the principal value branch before writing the final answer.
- Forgetting quadrant signs: Before choosing the angle, first check in which quadrants the trigonometric ratio is positive or negative.
- Directly writing answers without forming an equation: It is better to first assume something like \sin^{-1}x=y and then write \sin y=x. This makes the thinking process clear.
- Not understanding what inverse trigonometric functions mean: Expressions like \sin^{-1} or \cos^{-1} alone have no meaning unless they act on a value. For example, \sin^{-1}\left(\frac{1}{2}\right) asks for the angle whose sine is \frac{1}{2}.
- Choosing angles outside the principal value branch: Many angles may satisfy a trigonometric equation, but inverse trigonometric functions allow only the angle lying in the principal value interval.
- Mixing principal value intervals: Each inverse trigonometric function has its own principal value branch. For example, \sin^{-1}x and \cos^{-1}x do not have the same interval.
- Incorrect use of standard angles: Students often confuse values like \sin\frac{\pi}{6} and \cos\frac{\pi}{6}. Standard trigonometric values should be remembered carefully.
- Ignoring domain restrictions: Expressions like \sin^{-1}x and \cos^{-1}x are defined only for -1\le x\le 1.
Continue Learning
After completing Exercise 2.1 of Inverse Trigonometric Functions, you should now be comfortable with the concepts of principal value branches and evaluating inverse trigonometric functions using standard trigonometric values, quadrant signs and restricted intervals.
To strengthen your understanding further, make sure that you revise:
- Meaning and notation of inverse trigonometric functions
- Difference between inverse trigonometric functions and reciprocal trigonometric functions
- Principal value branches of all inverse trigonometric functions
- Standard trigonometric values and standard angles
- Signs of trigonometric functions in different quadrants
- Converting inverse trigonometric expressions into standard trigonometric equations
- Identifying the correct principal value among multiple possible angles
- Domain and range restrictions of inverse trigonometric functions
Explore More
Try creating your own examples involving inverse trigonometric functions and identify their principal values using standard angles, quadrant signs and principal value branches. This practice will strengthen conceptual understanding and improve confidence in solving inverse trigonometric function problems.
All the best and keep learning 👍
Discover more from Maths Better
Subscribe to get the latest posts sent to your email.
