Class 12 Maths · Original EZT practice
Inverse Trigonometric Functions (Class 12 Chapter 2): summary and practice
Sin maps many angles to the same value, so its inverse only works after we restrict the domain. That one idea - the principal value branch - decides every answer in this chapter. All questions are original EZT tutor practice following the chapter's own section order.
8 original worked examples, 8 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 20 October 2026:
1. Why the domain is restricted
An inverse exists only when the function is one-one and onto. Trigonometric functions repeat, so each is restricted to a branch where it passes the horizontal-line test; the inverse of that branch is the inverse trigonometric function.
2. Principal value branches
sin^-1 takes [-1, 1] to [-pi/2, pi/2]; cos^-1 takes [-1, 1] to [0, pi]; tan^-1 takes all real numbers to (-pi/2, pi/2). Every answer must land inside these ranges.
3. The two composition identities
sin(sin^-1 x) = x holds for every x in [-1, 1]. But sin^-1(sin x) = x holds only when x lies in the principal branch; outside it, you first fold the angle back into the branch.
4. Graphs as reflections
The graph of an inverse trigonometric function is the reflection of the restricted trigonometric graph in the line y = x - a quick visual check for domain and range.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Why the domain needs restricting
Choose the method: One input must give one output - so pick a one-one branch.
EZT practice question
Why is sin^-1 defined only on a restricted branch of sin?
Show the worked solution
- sin takes the same value at infinitely many angles, so it is not one-one on all of R.
- Restricting to [-pi/2, pi/2] makes it one-one and onto [-1, 1], so an inverse exists.
Answer: Because sin repeats; only on a one-one branch (the principal branch [-pi/2, pi/2]) does an inverse function exist.
Check: On that branch each value in [-1, 1] comes from exactly one angle.
Watch for this: Do not treat sin^-1 x as 1/sin x; the inverse function is not the reciprocal.
Now try it yourself
What is the domain of sin^-1 x?
Check your answer
[-1, 1].
It is the range of sin on the principal branch.
Quote the principal value branches
Choose the method: Memorise the table exactly as the chapter prints it.
EZT practice question
State the domain and principal value branch of sin^-1, cos^-1 and tan^-1.
Show the worked solution
- sin^-1: domain [-1, 1], range [-pi/2, pi/2].
- cos^-1: domain [-1, 1], range [0, pi].
- tan^-1: domain all real numbers, range (-pi/2, pi/2).
Answer: sin^-1 to [-pi/2, pi/2]; cos^-1 to [0, pi]; tan^-1 to (-pi/2, pi/2).
Check: cos^-1 uses a closed upper range while tan^-1 is open - pi/2 is never a tan^-1 value.
Watch for this: Do not give cos^-1 the same branch as sin^-1; its branch is [0, pi].
Now try it yourself
What is the range of tan^-1 x?
Check your answer
(-pi/2, pi/2).
tan maps that open interval onto all real numbers.
Evaluate sin^-1 values
Choose the method: Find the angle in [-pi/2, pi/2] whose sine matches.
EZT practice question
Find sin^-1(1/2) and sin^-1(-1).
Show the worked solution
- sin 30 degrees = 1/2, and 30 degrees is in the branch.
- sin(-90 degrees) = -1, and -90 degrees is the branch endpoint.
Answer: sin^-1(1/2) = pi/6; sin^-1(-1) = -pi/2.
Check: Both answers sit inside [-pi/2, pi/2] - required.
Watch for this: Do not answer 150 degrees for the first; it lies outside the branch even though sin 150 = 1/2.
Now try it yourself
Find sin^-1(-1/2).
Check your answer
-pi/6.
sin(-30 degrees) = -1/2 and -pi/6 is inside the branch.
Evaluate cos^-1 values, including negatives
Choose the method: Find the angle in [0, pi] whose cosine matches.
EZT practice question
Find cos^-1(-1/2) and cos^-1(0).
Show the worked solution
- cos 120 degrees = -1/2, and 120 degrees is inside [0, pi].
- cos 90 degrees = 0.
Answer: cos^-1(-1/2) = 2pi/3; cos^-1(0) = pi/2.
Check: Negative cosine lands in the second quadrant - exactly what the [0, pi] branch covers.
Watch for this: Do not answer -60 degrees for cos^-1(-1/2); the cos^-1 branch has no negative angles.
Now try it yourself
Find cos^-1(1).
Check your answer
0.
cos 0 = 1 and 0 is the branch endpoint.
Handle sin^-1(sin x) outside the branch
Choose the method: Fold the angle back into [-pi/2, pi/2] before answering.
EZT practice question
Evaluate sin^-1(sin 150 degrees).
Show the worked solution
- sin 150 = 1/2, so we want sin^-1(1/2).
- The branch answer is 30 degrees, not 150.
Answer: 30 degrees (pi/6) - 150 degrees lies outside the principal branch.
Check: sin 30 = sin 150 = 1/2, and 30 is the branch angle.
Watch for this: Do not write sin^-1(sin x) = x automatically; it holds only inside the branch.
Now try it yourself
Evaluate sin^-1(sin 120 degrees).
Check your answer
60 degrees.
sin 120 = sin 60, and 60 degrees is inside the branch.
Handle cos^-1(cos x) with a negative angle
Choose the method: Use evenness or fold into [0, pi].
EZT practice question
Evaluate cos^-1(cos(-60 degrees)).
Show the worked solution
- cos(-60) = cos 60, since cosine is even.
- cos^-1(cos 60) = 60 degrees, inside the branch.
Answer: 60 degrees (pi/3).
Check: The cos^-1 branch [0, pi] has no negative angles, so -60 degrees could not be the answer.
Watch for this: Do not return -60 degrees; fold it to the branch first.
Now try it yourself
Evaluate cos^-1(cos 390 degrees).
Check your answer
30 degrees.
cos 390 = cos 30, and 30 degrees is inside [0, pi].
Compose different inverses
Choose the method: Name the angle, draw the right triangle, read the ratio.
EZT practice question
Evaluate sin(cos^-1(3/5)).
Show the worked solution
- Let theta = cos^-1(3/5), so cos theta = 3/5.
- Then sin theta = sqrt(1 - 9/25) = 4/5, positive because the branch sits in [0, pi] where sin is non-negative.
Answer: 4/5.
Check: A 3-4-5 triangle: adjacent 3, hypotenuse 5, opposite 4.
Watch for this: Do not write sin theta = plus or minus 4/5; the branch fixes the sign.
Now try it yourself
Evaluate cos(sin^-1(5/13)).
Check your answer
12/13.
A 5-12-13 triangle: sin gives the opposite, cos needs the adjacent.
Apply the two identities correctly
Choose the method: sin(sin^-1 x) is unconditional on [-1, 1]; sin^-1(sin x) is not.
EZT practice question
For which x does sin(sin^-1 x) = x hold, and for which x does sin^-1(sin x) = x hold?
Show the worked solution
- sin(sin^-1 x) = x for every x in [-1, 1] - the domain of sin^-1.
- sin^-1(sin x) = x only for x in [-pi/2, pi/2] - the principal branch.
Answer: The first holds on [-1, 1]; the second only on [-pi/2, pi/2].
Check: Outside the branch, sin^-1(sin x) returns the folded-back angle instead.
Watch for this: Do not claim either identity for all real x; both have limited domains of validity.
Now try it yourself
Is sin^-1(sin 3pi/4) equal to 3pi/4?
Check your answer
No - it is pi/4.
3pi/4 lies outside [-pi/2, pi/2]; sin 3pi/4 = sin pi/4.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Principal value branch of sin^-1?
Show answer and reason
[-pi/2, pi/2].
Every sin^-1 answer must land inside it.
2. Principal value branch of cos^-1?
Show answer and reason
[0, pi].
It covers the second quadrant, where cosine is negative.
3. Principal value branch of tan^-1?
Show answer and reason
(-pi/2, pi/2).
Open at the ends - tan blows up at pi/2.
4. Is sin^-1 x the same as 1/sin x?
Show answer and reason
No. The inverse function undoes sin; the reciprocal is cosec x.
Mixing them breaks every problem in the chapter.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Find sin^-1(-1/2).
- Find cos^-1(0).
- Evaluate sin^-1(sin 120 degrees).
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Questions about this practice page
- Why is the domain of sin restricted before defining sin^-1?
- Sin repeats its values, so it is not one-one on all real numbers. Restricting to [-pi/2, pi/2] makes it one-one and onto [-1, 1], which an inverse function requires.
- What is a principal value?
- The unique answer of an inverse trigonometric function taken from its principal value branch - for example sin^-1(1/2) is pi/6, not 5pi/6.
- How do I find sin^-1(sin x) when x is outside the branch?
- Use symmetries to fold x into [-pi/2, pi/2]: find the branch angle with the same sine, like sin^-1(sin 150 degrees) = 30 degrees.
- How are the graphs of sin and sin^-1 related?
- The sin^-1 graph is the reflection of the restricted sin graph in the line y = x.
- Are these the NCERT exercise questions for Inverse Trigonometric Functions (Class 12)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.