MathsCBSE Class 12 · Inverse Trigonometric Functions

Composite Inverse Trigonometric Expressions: sin⁻¹(sin x) and More (Class 12)

Q: Why is sin inverse of sin(3π/4) not equal to 3π/4?

Sin inverse always returns an angle in [−π/2, π/2]. Since 3π/4 is outside this interval, the answer is the angle inside it with the same sine, which is π/4.

Q: Is sin inverse of sin x = x a property I can use?

Only for x in [−π/2, π/2], and this is one of the two results the 2026-27 syllabus keeps (along with sin(sin inverse x) = x for x in [−1, 1]). Outside that interval you must reduce the angle into the branch.

Q: How do I find cos(tan inverse x) quickly?

Let θ = tan inverse x, so tan θ = x and θ is between −π/2 and π/2, where cosine is positive. A right triangle with opposite side x and adjacent side 1 has hypotenuse √(1 + x²), so cos θ = 1/√(1 + x²).

Q: Which NCERT section covers these evaluations?

Section 2.2 of Chapter 2, Inverse Trigonometric Functions, in NCERT Mathematics Part I for Class 12, through the principal value branches. The properties section was dropped in the rationalised book.

Q: What if the angle is given as a plain number like 4?

It is in radians. Compare it with π/2 ≈ 1.57, π ≈ 3.14, 3π/2 ≈ 4.71 and 2π ≈ 6.28, then use the matching reduction, such as π − 4 for sine.

Updated 15 October 2026 · Fact-checked

sin(sin⁻¹x) = x for every x in [−1, 1], but sin⁻¹(sin x) = x only when x lies in [−π/2, π/2]. For any other angle, find the angle inside the principal branch that has the same sine (or cosine, or tangent). The same branch idea gives cos⁻¹(cos x) and tan⁻¹(tan x).

sin⁻¹(sin x) only returns x inside the green branch; for 3π/4, find the branch angle π/4 with the same sine.

Illustration drawn for QuestionBuddy.

Understand Composite Inverse Trigonometric Expressions

A composite expression applies a trigonometric function and an inverse trigonometric function one after the other. There are two orders, and they behave very differently.

**Order 1: trig of inverse, such as $sin (sin^{- 1} x)$ .** Here $sin^{- 1} x$ is an angle whose sine is $x$ , so taking its sine gives back $x$ . This works for **every $x$ in the domain** $[- 1, 1]$ . Likewise $cos (cos^{- 1} x) = x$ for $x \in [- 1, 1]$ and $tan (tan^{- 1} x) = x$ for all real $x$ . Outside the domain, the expression is not defined at all: $sin (sin^{- 1} 2)$ is meaningless.

**Order 2: inverse of trig, such as $sin^{- 1} (sin x)$ .** The answer must be a principal value, so it always lies in the branch $[- \frac{π}{2}, \frac{π}{2}]$ . Hence $sin^{- 1} (sin x) = x$ **only when $x \in [- \frac{π}{2}, \frac{π}{2}]$ **. This is the one inverse-function fact the 2026-27 syllabus keeps from the dropped properties section. For any other angle, find the angle $θ$ in the branch with $sin θ = sin x$ , using the standard reductions:

$sin (π - x) = sin x$ , $sin (x - 2 π) = sin x$ , $sin (- x) = - sin x$ ;
$cos (2 π - x) = cos x$ , $cos (- x) = cos x$ ;
$tan (x - π) = tan x$ (tangent repeats every $π$ ).

For example, $\frac{3 π}{4}$ is outside $[- \frac{π}{2}, \frac{π}{2}]$ , but $sin \frac{3 π}{4} = sin (π - \frac{3 π}{4}) = sin \frac{π}{4}$ , so $sin^{- 1} (sin \frac{3 π}{4}) = \frac{π}{4}$ . For cosine, $\frac{3 π}{4}$ is already in $[0, π]$ , so $cos^{- 1} (cos \frac{3 π}{4}) = \frac{3 π}{4}$ . For tangent, $tan \frac{3 π}{4} = tan (\frac{3 π}{4} - π) = tan (- \frac{π}{4})$ , so $tan^{- 1} (tan \frac{3 π}{4}) = - \frac{π}{4}$ . The same angle gives three different answers, because the three branches differ.

A safe two-step method: first compute the inner value exactly ( $sin \frac{3 π}{4} = \frac{1}{2}$ ), then find its principal value ( $sin^{- 1} \frac{1}{2} = \frac{π}{4}$ ). For angles without a table value, such as $sin^{- 1} (sin 4)$ with 4 in radians, use the reductions instead: $4$ lies between $\frac{π}{2}$ and $\frac{3 π}{2}$ , so $π - 4$ lies in the branch and has the same sine; the answer is $π - 4$ .

**Mixed functions, such as $sin (cos^{- 1} \frac{3}{5})$ .** Let $θ = cos^{- 1} \frac{3}{5}$ . Then $cos θ = \frac{3}{5}$ and $θ \in [0, π]$ , where sine is never negative. So $sin θ = 1 - \frac{9}{25} = \frac{4}{5}$ . The branch decides the sign: in $cos^{- 1}$ problems sine is $\geq 0$ ; in $sin^{- 1}$ and $tan^{- 1}$ problems cosine is $\geq 0$ . A right-triangle sketch gives the size, the branch gives the sign.

Simplifying with a variable. Expressions like $tan^{- 1} \frac{1 - c o s x}{1 + c o s x}$ become $tan^{- 1} (tan \frac{x}{2})$ , which equals $\frac{x}{2}$ only after you check that $\frac{x}{2}$ lies in the branch for the given interval of $x$ . This is NCERT Section 2.2 material; keep to the branch reasoning and avoid the dropped identities.

Composite Inverse Trigonometric Expressions: formulas to remember

Trig of inverse: $sin (sin^{- 1} x) = x, cos (cos^{- 1} x) = x (x \in [- 1, 1]); tan (tan^{- 1} x) = x (x \in R)$; Valid on the whole domain of the inverse function.
Inverse sine of sine: $sin^{- 1} (sin x) = x ⟺ x \in [- \frac{π}{2}, \frac{π}{2}]$; For $x \in [\frac{π}{2}, \frac{3 π}{2}]$ the value is $π - x$ .
Inverse cosine of cosine: $cos^{- 1} (cos x) = x ⟺ x \in [0, π]$; For $x \in [π, 2 π]$ the value is $2 π - x$ ; for $x \in [- π, 0]$ it is $- x$ .
Inverse tangent of tangent: $tan^{- 1} (tan x) = x ⟺ x \in (- \frac{π}{2}, \frac{π}{2})$; For $x \in (\frac{π}{2}, \frac{3 π}{2})$ the value is $x - π$ .
Sine of an inverse cosine: $sin (cos^{- 1} x) = 1 - x^{2}$; Never negative, because $cos^{- 1} x \in [0, π]$ .
Cosine of an inverse sine or tangent: $cos (sin^{- 1} x) = 1 - x^{2}, cos (tan^{- 1} x) = \frac{1}{1 + x ^{2}}$; Positive, because these branches lie in $[- \frac{π}{2}, \frac{π}{2}]$ .

How to solve Composite Inverse Trigonometric Expressions questions

Use this routine for $f^{- 1} (f (x))$ , $f (g^{- 1} (x))$ and simplification questions.

1Identify the order: trig of inverse (usually just returns $x$ ) or inverse of trig (needs the branch).
2For inverse of trig with a standard angle, compute the inner value exactly, then take its principal value.
3For a non-standard angle (or a variable), check whether it already lies in the branch. If not, move it into the branch with a reduction that keeps the function's value: $π - x$ for sine, $2 π - x$ or $- x$ for cosine, $x \pm π$ for tangent; subtract multiples of $2 π$ first if the angle is large.
4Confirm the new angle is in the branch, for example $- \frac{π}{2} \leq π - 4 \leq \frac{π}{2}$ .
5For mixed expressions like $tan (cos^{- 1} x)$ , let $θ$ be the inverse value, write the known ratio, find the missing side of a right triangle, and fix the sign from the branch of $θ$ .
6For a simplification with a variable, rewrite the inside as a single trigonometric function of one angle, then check the angle stays in the branch over the given interval before cancelling.
7State the final value; give decimals only if asked.

Composite Inverse Trigonometric Expressions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

sin⁻¹(sin x) loses the quadrant

Suppose a Ferris wheel app finds a cabin's position angle from the horizontal using only the cabin's height above the hub: it computes $θ = sin^{- 1} (\frac{h}{r})$ , where $r$ is the radius. A cabin is actually at the angle $\frac{5 π}{6}$ , on the left side of the wheel.

What angle does the app report? How can the true angle be recovered if the app also knows the cabin is on the left (its horizontal position is negative)?

Try it first, then show the solutionSolution

Height ratio $\frac{h}{r} = sin \frac{5 π}{6} = \frac{1}{2}$ .
The app reports $sin^{- 1} \frac{1}{2} = \frac{π}{6}$ , which is $sin^{- 1} (sin \frac{5 π}{6})$ , not $\frac{5 π}{6}$ .
$\frac{π}{6}$ and $\frac{5 π}{6}$ have the same sine; the left-side cabin is the one at $π - \frac{π}{6} = \frac{5 π}{6}$ .
So the app should report $π - θ$ whenever the horizontal position is negative.

Answer: The app reports $\frac{π}{6}$ ; the true angle is $π - \frac{π}{6} = \frac{5 π}{6}$ .

Memory hook: Height alone cannot tell left from right: $sin^{- 1} (sin x)$ always lands on the right half.

tan⁻¹(tan x) and the direction of a point

Imagine a drone's controller computes the direction of a target from the drone using $θ = tan^{- 1} (\frac{y}{x})$ . The target is at $(- 1, - 3)$ relative to the drone (west and south of it).

What direction does the formula give? What is the true direction, measured from the positive x-axis and taken between $- π$ and $π$ ?

Try it first, then show the solutionSolution

$\frac{y}{x} = \frac{- 3}{- 1} = 3$ , so the formula gives $tan^{- 1} 3 = \frac{π}{3}$ , pointing north-east.
But the point is in the third quadrant, so the true angle $α$ satisfies $tan α = 3$ with both coordinates negative.
$tan$ repeats every $π$ : $α = \frac{π}{3} - π = - \frac{2 π}{3}$ .
Check: $cos (- \frac{2 π}{3}) = - \frac{1}{2}$ and $sin (- \frac{2 π}{3}) = - \frac{3}{2}$ , matching $(- 1, - 3)$ scaled by 2.

Answer: The formula gives $\frac{π}{3}$ ; the true direction is $- \frac{2 π}{3}$ (south-west).

Memory hook: $tan^{- 1}$ only looks right, between $- \frac{π}{2}$ and $\frac{π}{2}$ ; points on the left need a shift of $π$ .

Tricks to remember Composite Inverse Trigonometric Expressions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Outside-in works, inside-out needs a passport
$sin (sin^{- 1} x)$ always returns $x$ . $sin^{- 1} (sin x)$ returns $x$ only if $x$ holds a branch passport, $[- \frac{π}{2}, \frac{π}{2}]$ .
Use it when: Spotting which composite needs work.
Mirror for sine, flip for cosine, slide for tangent
Sine: mirror in $\frac{π}{2}$ , use $π - x$ . Cosine: flip in $π$ , use $2 π - x$ (or $- x$ ). Tangent: slide by $π$ , use $x - π$ .
Use it when: Bringing an angle into the branch without changing the function value.
Size from the triangle, sign from the branch
For $sin (cos^{- 1} x)$ or $cos (tan^{- 1} x)$ , the right triangle gives the number; the branch of the inverse function gives the sign.
Use it when: Mixed-function evaluations.
Radians are not degrees
In $sin^{- 1} (sin 4)$ , 4 means 4 radians, about $229°$ . Compare with $\frac{π}{2} \approx 1.57$ , $π \approx 3.14$ and $\frac{3 π}{2} \approx 4.71$ .
Use it when: HOTS questions with whole-number angles.

Composite Inverse Trigonometric Expressions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Same idea, two branches · 2 marks · Very short answerBoard level

Evaluate $sin^{- 1} (sin \frac{2 π}{3}) + cos^{- 1} (cos \frac{5 π}{4})$ .

Try it first, then show the solutionSolution

$\frac{2 π}{3} \in / [- \frac{π}{2}, \frac{π}{2}]$ . $sin \frac{2 π}{3} = sin (π - \frac{2 π}{3}) = sin \frac{π}{3}$ , and $\frac{π}{3}$ is in the branch, so the first term is $\frac{π}{3}$ . (1 mark)
$\frac{5 π}{4} \in / [0, π]$ . $cos \frac{5 π}{4} = cos (2 π - \frac{5 π}{4}) = cos \frac{3 π}{4}$ , and $\frac{3 π}{4} \in [0, π]$ , so the second term is $\frac{3 π}{4}$ . (½ mark)
Sum $= \frac{π}{3} + \frac{3 π}{4} = \frac{13 π}{12}$ . (½ mark)

Answer: $\frac{13 π}{12}$

Example 2: Mixed functions with a sign decided by the branch · 2 marks · Very short answerBoard level

Find (i) $sin (cos^{- 1} \frac{3}{5})$ and (ii) $tan (cos^{- 1} (- \frac{3}{5}))$ .

Try it first, then show the solutionSolution

(i) Let $θ = cos^{- 1} \frac{3}{5}$ , so $cos θ = \frac{3}{5}$ , $θ \in [0, π]$ and $sin θ \geq 0$ . Hence $sin θ = 1 - \frac{9}{25} = \frac{4}{5}$ . (1 mark)
(ii) Let $φ = cos^{- 1} (- \frac{3}{5})$ , so $cos φ = - \frac{3}{5}$ with $φ \in (\frac{π}{2}, π)$ . Then $sin φ = \frac{4}{5}$ and $tan φ = \frac{4/5}{- 3/5} = - \frac{4}{3}$ . (1 mark)

Answer: (i) $\frac{4}{5}$ (ii) $- \frac{4}{3}$

Example 3: A large angle inside a different function · 2 marks · Very short answerHarder board level

Find the value of $sin^{- 1} (cos \frac{29 π}{7})$ .

Try it first, then show the solutionSolution

Remove whole turns: $\frac{29 π}{7} = 4 π + \frac{π}{7}$ , so $cos \frac{29 π}{7} = cos \frac{π}{7}$ . (½ mark)
Convert to sine: $cos \frac{π}{7} = sin (\frac{π}{2} - \frac{π}{7}) = sin \frac{5 π}{14}$ . (½ mark)
$\frac{5 π}{14} \in [- \frac{π}{2}, \frac{π}{2}]$ , so $sin^{- 1} (sin \frac{5 π}{14}) = \frac{5 π}{14}$ . (1 mark)

Answer: $\frac{5 π}{14}$

Example 4: Simplifying with a variable · 3 marks · Short answerHarder board level

Simplify $tan^{- 1} \frac{1 - c o s x}{1 + c o s x}$ , where $0 < x < π$ .

Try it first, then show the solutionSolution

Use $1 - cos x = 2 sin^{2} \frac{x}{2}$ and $1 + cos x = 2 cos^{2} \frac{x}{2}$ . (1 mark)
So $\frac{1 - c o s x}{1 + c o s x} = tan^{2} \frac{x}{2} = tan \frac{x}{2} = tan \frac{x}{2}$ , since $0 < \frac{x}{2} < \frac{π}{2}$ makes it positive. (1 mark)
The expression is $tan^{- 1} (tan \frac{x}{2})$ , and $\frac{x}{2} \in (0, \frac{π}{2}) \subset (- \frac{π}{2}, \frac{π}{2})$ , so it equals $\frac{x}{2}$ . (1 mark)

Answer: $\frac{x}{2}$

Example 5: Angles in radians that are not multiples of π · 3 marks · Short answerHOTS / competency

Find the value of $sin^{- 1} (sin 4) + cos^{- 1} (cos 4) + tan^{- 1} (tan 4)$ , the angles being in radians.

Try it first, then show the solutionSolution

$π \approx 3.14 < 4 < \frac{3 π}{2} \approx 4.71$ . (½ mark)
Sine: $sin 4 = sin (π - 4)$ and $π - 4 \approx - 0.86 \in [- \frac{π}{2}, \frac{π}{2}]$ , so $sin^{- 1} (sin 4) = π - 4$ . (1 mark)
Cosine: $cos 4 = cos (2 π - 4)$ and $2 π - 4 \approx 2.28 \in [0, π]$ , so $cos^{- 1} (cos 4) = 2 π - 4$ . (½ mark)
Tangent: $tan 4 = tan (4 - π)$ and $4 - π \approx 0.86 \in (- \frac{π}{2}, \frac{π}{2})$ , so $tan^{- 1} (tan 4) = 4 - π$ . (½ mark)
Sum $= (π - 4) + (2 π - 4) + (4 - π) = 2 π - 4$ . (½ mark)

Answer: $2 π - 4$

Example 6: Case study: the zigzag graph of sin⁻¹(sin x) · 4 marks · Case-basedHOTS / competency

Suppose a student plots $y = sin^{- 1} (sin x)$ for $0 \leq x \leq 2 π$ on a graphing app and sees straight-line pieces.
(i) Find $y$ at $x = \frac{π}{3}$ . (1 mark)
(ii) Find $y$ at $x = \frac{5 π}{6}$ . (1 mark)
(iii) Find a formula for $y$ when $\frac{π}{2} \leq x \leq \frac{3 π}{2}$ . (1 mark)
(iv) Find a formula for $y$ when $\frac{3 π}{2} \leq x \leq 2 π$ . (1 mark)

Try it first, then show the solutionSolution

(i) $\frac{π}{3}$ is in the branch, so $y = \frac{π}{3}$ . (1 mark)
(ii) $sin \frac{5 π}{6} = sin \frac{π}{6}$ , so $y = \frac{π}{6}$ . (1 mark)
(iii) For $\frac{π}{2} \leq x \leq \frac{3 π}{2}$ , $π - x \in [- \frac{π}{2}, \frac{π}{2}]$ and $sin (π - x) = sin x$ , so $y = π - x$ . (1 mark)
(iv) For $\frac{3 π}{2} \leq x \leq 2 π$ , $x - 2 π \in [- \frac{π}{2}, 0]$ and $sin (x - 2 π) = sin x$ , so $y = x - 2 π$ . (1 mark)

Answer: (i) $\frac{π}{3}$ (ii) $\frac{π}{6}$ (iii) $y = π - x$ (iv) $y = x - 2 π$

Where students lose marks in Composite Inverse Trigonometric Expressions

The slips examiners see most often, and what to write instead.

Mistake: Cancelling sin⁻¹ and sin automatically
$sin^{- 1} (sin x) = x$ is true only inside the branch.
Wrong working
$sin^{- 1} (sin \frac{3 π}{4}) = \frac{3 π}{4}$ .
Right working
$sin \frac{3 π}{4} = \frac{1}{2}$ , so $sin^{- 1} (sin \frac{3 π}{4}) = \frac{π}{4}$ .
Do this: Before cancelling, write 'since $x \in$ (branch)'. If you cannot, reduce the angle.
Usually costs: The value (1 to 2 marks).
Mistake: Using the sine reduction for cosine
$π - x$ keeps the sine but changes the sign of the cosine.
Wrong working
$cos^{- 1} (cos \frac{5 π}{4}) = π - \frac{5 π}{4} = - \frac{π}{4}$ .
Right working
$cos \frac{5 π}{4} = cos (2 π - \frac{5 π}{4}) = cos \frac{3 π}{4}$ , so the value is $\frac{3 π}{4}$ .
Do this: Learn the matching reductions: sine $π - x$ , cosine $2 π - x$ or $- x$ , tangent $x - π$ .
Usually costs: The value; a negative answer for $\cos^{-1}$ is a giveaway.
Mistake: Wrong sign in mixed expressions
The triangle gives only the size; the branch gives the sign.
Wrong working
$tan (cos^{- 1} (- \frac{3}{5})) = \frac{4}{3}$ .
Right working
The angle is in $(\frac{π}{2}, π)$ where tangent is negative: $- \frac{4}{3}$ .
Do this: Write the interval of the angle before finding any ratio.
Usually costs: 1 mark.
Mistake: Dropping the modulus after a square root
$tan^{2} \frac{x}{2} = tan \frac{x}{2}$ , which equals $tan \frac{x}{2}$ only when it is non-negative.
Wrong working
$tan^{2} \frac{x}{2} = tan \frac{x}{2}$ for all $x$ .
Right working
For $0 < x < π$ , $tan \frac{x}{2} > 0$ , so the modulus can be removed.
Do this: Use the given interval to justify every removal of a modulus.
Usually costs: Half to one mark in simplification questions.

Exam tips

Evaluations of the type $sin^{- 1} (sin a) + cos^{- 1} (cos b)$ were asked as 2-mark questions in the 2023 and 2026 board papers; show the reduced angle and say it lies in the branch.
Treat each term separately: the same angle can give different results under $sin^{- 1}$ , $cos^{- 1}$ and $tan^{- 1}$ .
For large angles like $\frac{33 π}{5}$ , remove multiples of $2 π$ first, then reduce.
In mixed expressions, a quick right-triangle sketch with the sides labelled earns method marks even if the arithmetic slips.
Avoid quoting identities such as $tan^{- 1} x + tan^{- 1} y$ ; they are out of the 2026-27 syllabus and branch reasoning is always enough.

Practice questions on Composite Inverse Trigonometric Expressions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The value of $sin^{- 1} (sin \frac{5 π}{6})$ is:
(a) $\frac{5 π}{6}$
(b) $\frac{π}{6}$
(c) $- \frac{π}{6}$
(d) $\frac{7 π}{6}$
Hint
Is $\frac{5 π}{6}$ in $[- \frac{π}{2}, \frac{π}{2}]$ ?
Answer
(b) $\frac{π}{6}$ , since $sin \frac{5 π}{6} = sin \frac{π}{6}$ and $\frac{π}{6}$ is in the branch.
Q2 · 1 mark · MCQBoard level
The value of $cos^{- 1} (cos (- \frac{π}{3}))$ is:
(a) $- \frac{π}{3}$
(b) $\frac{π}{3}$
(c) $\frac{2 π}{3}$
(d) $\frac{5 π}{3}$
Hint
Cosine is even; the answer must lie in $[0, π]$ .
Answer
(b) $\frac{π}{3}$ .
Q3 · 2 marks · Very short answerBoard level
Find $k$ if $tan^{- 1} [k cos (2 sin^{- 1} \frac{1}{2})] = \frac{π}{3}$ .
Hint
Evaluate from the inside: $sin^{- 1} \frac{1}{2}$ first.
Answer
$sin^{- 1} \frac{1}{2} = \frac{π}{6}$ , so $cos \frac{π}{3} = \frac{1}{2}$ . Then $tan^{- 1} \frac{k}{2} = \frac{π}{3}$ gives $\frac{k}{2} = 3$ , so $k = 23$ .
Q4 · 2 marks · Very short answerHarder board level
Find the value of $tan^{- 1} (tan \frac{3 π}{5})$ .
Hint
Slide by $π$ into $(- \frac{π}{2}, \frac{π}{2})$ .
Answer
$tan \frac{3 π}{5} = tan (\frac{3 π}{5} - π) = tan (- \frac{2 π}{5})$ , and $- \frac{2 π}{5}$ is in the branch, so the value is $- \frac{2 π}{5}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $sin^{- 1} (sin \frac{2 π}{3}) = \frac{2 π}{3}$ .
Reason (R): $sin^{- 1} (sin x) = x$ holds only for $x \in [- \frac{π}{2}, \frac{π}{2}]$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Is $\frac{2 π}{3}$ in the branch of $sin^{- 1}$ ?
Answer
(d). $\frac{2 π}{3}$ is outside the branch, so $sin^{- 1} (sin \frac{2 π}{3}) = \frac{π}{3}$ and A is false; R is true.
Q6 · 3 marks · Short answerHarder board level
Evaluate (i) $sec (tan^{- 1} (- \frac{5}{12}))$ and (ii) $cot (sin^{- 1} (- \frac{3}{5}))$ .
Hint
Both angles lie in $(- \frac{π}{2}, 0)$ , where cosine is positive and sine is negative.
Answer
(i) $θ = tan^{- 1} (- \frac{5}{12}) \in (- \frac{π}{2}, 0)$ : hypotenuse 13, $cos θ = \frac{12}{13}$ , so $sec θ = \frac{13}{12}$ . (ii) $φ = sin^{- 1} (- \frac{3}{5}) \in [- \frac{π}{2}, 0)$ : $cos φ = \frac{4}{5}$ , $sin φ = - \frac{3}{5}$ , so $cot φ = - \frac{4}{3}$ .
Q7 · 3 marks · Short answerHOTS / competency
(i) Show that $cos^{- 1} (cos x) = 2 π - x$ for $π \leq x \leq 2 π$ . (ii) Hence find $cos^{- 1} (cos 5)$ and also $sin^{- 1} (sin 5)$ , the angles being in radians.
Hint
$\frac{3 π}{2} \approx 4.71 < 5 < 2 π \approx 6.28$ .
Answer
(i) If $π \leq x \leq 2 π$ then $0 \leq 2 π - x \leq π$ and $cos (2 π - x) = cos x$ , so $cos^{- 1} (cos x) = 2 π - x$ . (ii) $cos^{- 1} (cos 5) = 2 π - 5$ . For sine, $sin 5 = sin (5 - 2 π)$ and $5 - 2 π \approx - 1.28 \in [- \frac{π}{2}, \frac{π}{2}]$ , so $sin^{- 1} (sin 5) = 5 - 2 π$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.For which $x$ is $sin^{- 1} (sin x) = x$ ?Show answer
$x \in [- \frac{π}{2}, \frac{π}{2}]$ .
2.For which $x$ is $sin (sin^{- 1} x) = x$ ?Show answer
Every $x \in [- 1, 1]$ .
3. $cos^{- 1} (cos \frac{7 π}{6}) = ?$ Show answer
$\frac{5 π}{6}$ (use $2 π - \frac{7 π}{6}$ ).
4. $tan^{- 1} (tan \frac{3 π}{4}) = ?$ Show answer
$- \frac{π}{4}$ .
5. $sin (cos^{- 1} \frac{5}{13}) = ?$ Show answer
$\frac{12}{13}$ .
6. $sin^{- 1} (sin 2)$ with 2 in radians?Show answer
$π - 2$ , since $\frac{π}{2} < 2 < π$ .

Composite Inverse Trigonometric Expressions: frequently asked questions

Why is sin inverse of sin(3π/4) not equal to 3π/4?

Sin inverse always returns an angle in [−π/2, π/2]. Since 3π/4 is outside this interval, the answer is the angle inside it with the same sine, which is π/4.

Is sin inverse of sin x = x a property I can use?

Only for x in [−π/2, π/2], and this is one of the two results the 2026-27 syllabus keeps (along with sin(sin inverse x) = x for x in [−1, 1]). Outside that interval you must reduce the angle into the branch.

How do I find cos(tan inverse x) quickly?

Let θ = tan inverse x, so tan θ = x and θ is between −π/2 and π/2, where cosine is positive. A right triangle with opposite side x and adjacent side 1 has hypotenuse √(1 + x²), so cos θ = 1/√(1 + x²).

Which NCERT section covers these evaluations?

Section 2.2 of Chapter 2, Inverse Trigonometric Functions, in NCERT Mathematics Part I for Class 12, through the principal value branches. The properties section was dropped in the rationalised book.

What if the angle is given as a plain number like 4?

It is in radians. Compare it with π/2 ≈ 1.57, π ≈ 3.14, 3π/2 ≈ 4.71 and 2π ≈ 6.28, then use the matching reduction, such as π − 4 for sine.