Inverse Trigonometric Functions | ISC Class 12 Maths Notes

This note covers inverse trigonometric functions, domain and range, principal value branches, graphs, exact values, compositions, reciprocal and complementary relations, sums and differences, double-angle and triple-angle formulae, and equations involving inverse trigonometric functions.

What is an inverse trigonometric function?

A function assigns exactly one output to each permitted input. Its domain is the set of permitted inputs; its range is the set of outputs obtained. The codomain is the specified set into which the function maps.

A function is one-one when different inputs have different outputs. It is onto when every member of its codomain is obtained. A function that is both is called bijective, and has an inverse that reverses its input-output correspondence.

Write f for a function and f⁻¹ for its inverse. If f takes an input x to an output y, its inverse takes y back to x. Thus the inverse exchanges the domain and range of the original bijective function.

Why must the trigonometric domain be restricted?

The symbols sin, cos, tan, cot, sec and cosec denote sine, cosine, tangent, cotangent, secant and cosecant. These trigonometric functions repeat values on their natural domains. Reversing the correspondence without choosing a suitable domain would allow several angle outputs for the same input.

A restriction keeps only a selected part of a function's domain. Restricting sine to angles from −π/2 to π/2 makes it one-one with range from −1 to 1. Here π is the circle constant. All angles are measured in radians: an angle in radians is the arc length divided by the circle radius. A straight angle measures π radians.

Definition: An inverse trigonometric function returns the angle in a chosen branch whose trigonometric value equals the given input. A branch is an inverse obtained from a suitable restricted domain of the original trigonometric function.

For example, y = sin⁻¹x means sin y = x, with y restricted to the chosen branch. The notation is also read as arc sine. The analogous meanings apply to the other five inverse trigonometric functions.

Note: sin⁻¹x is an inverse-function value, whereas (sin x)⁻¹ is the reciprocal 1/sin x. A reciprocal is one divided by a non-zero quantity. Parentheses distinguish these two meanings.

What are the domains and principal value ranges?

The principal value branch is the conventional branch used when no other branch is specified. The angle returned within its range is the principal value. Knowing this range is part of knowing the inverse function itself.

In the table, R denotes the set of real numbers. Square brackets include endpoints; round brackets exclude them. A minus sign between sets means removal: R − (−1, 1) removes all real numbers strictly between −1 and 1. Braces enclose specified individual elements.

The symbols ≤ and ≥ mean “less than or equal to” and “greater than or equal to”. The strict inequalities < and > mean “less than” and “greater than”. Equality is written =.

How should the endpoints be read?

For sine and cosine inverses, both −1 and 1 are admissible inputs. For cosecant and secant inverses, those same endpoints are included, but every input strictly between them is excluded. An excluded input gives no real value for that inverse function.

The tangent inverse can take every real input, but its output cannot equal either −π/2 or π/2. The cotangent inverse also accepts every real input, while its output cannot equal 0 or π. These exclusions follow from the corresponding original functions being undefined at those angles.

Keep the two columns separate: the domain checks the number supplied, while the range checks the angle obtained. Restricting the original trigonometric domain creates the inverse range. The original range becomes the inverse domain.

The omitted angle 0 in the cosecant inverse range and the omitted angle π/2 in the secant inverse range are essential. Neither omission removes −1 or 1 from the corresponding input domain.

How are principal values of sine and cosine inverses found?

Let θ, the Greek letter theta, denote the angle being sought. To find sin⁻¹x, solve sin θ = x and choose θ in [−π/2, π/2]. To find cos⁻¹x, solve cos θ = x and choose θ in [0, π].

The reference angle is the acute angle used to recognise the magnitude of a trigonometric value. It helps identify candidates, but the principal range selects the answer. A negative cosine input requires an angle greater than π/2 and at most π, rather than a negative angle.

What steps keep the branch visible?

1. Check that the input belongs to [−1, 1], so that the requested real inverse exists.
2. Name the unknown angle and write the corresponding sine or cosine equation.
3. Identify an angle with the required trigonometric value, using exact special-angle values.
4. Check the principal range before recording the inverse-function value as the answer.

Worked example 1. Find the principal value of sin⁻¹(1/√2). The square-root sign √ denotes the non-negative square root.

Answer: Put θ = sin⁻¹(1/√2). Then sin θ = 1/√2. Since sin(π/4) = 1/√2 and π/4 belongs to [−π/2, π/2], the principal value is π/4.

Worked example 2. Find sin⁻¹(−1/2).

Answer: The angle −π/6 has sine −1/2 and lies in [−π/2, π/2]. Therefore sin⁻¹(−1/2) = −π/6. Choosing a positive angle with the same sine would fail the required principal-range check.

Worked example 3. Find cos⁻¹(−1/2).

Answer: Since cos(2π/3) = −1/2 and 2π/3 lies in [0, π], cos⁻¹(−1/2) = 2π/3. The negative sign in the input does not make the inverse-cosine output negative.

These examples illustrate two different branch choices. Inverse sine can return negative angles, but inverse cosine returns angles in [0, π]. A familiar trigonometric value alone is insufficient unless the selected angle also satisfies the relevant interval condition.

How do inverse tangent and inverse cotangent differ?

The inverse tangent returns an angle strictly between −π/2 and π/2. The inverse cotangent returns an angle strictly between 0 and π. Although both accept every real input, their principal ranges are different, so their signs cannot be treated in the same way.

For a negative input, the principal tangent angle is negative. The principal cotangent angle lies between π/2 and π. For a positive input, both angles lie between 0 and π/2, but they are generally different angles.

How does the range determine the answer?

Worked example 4. Find tan⁻¹(−√3).

Answer: Let θ = tan⁻¹(−√3). Then tan θ = −√3 and −π/2 < θ < π/2. Since tan(−π/3) = −√3, the required principal value is −π/3.

Worked example 5. Find cot⁻¹(−1/√3).

Answer: The reference angle is π/3, whose cotangent is 1/√3. The angle 2π/3 has cotangent −1/√3 and belongs to (0, π). Hence cot⁻¹(−1/√3) = 2π/3.

Cotangent is the reciprocal of tangent wherever both expressions are defined. This reciprocal relation between the original trigonometric functions does not permit an unconditional replacement of cot⁻¹x by tan⁻¹(1/x). The resulting angle must still belong to the cotangent principal range.

For x > 0, cot⁻¹x = tan⁻¹(1/x). For x < 0, cot⁻¹x = π + tan⁻¹(1/x). The added π in the second case moves the negative tangent-inverse angle into the cotangent principal range.

At an input of zero, the inverse tangent gives 0, while the inverse cotangent gives π/2. Neither inverse requires division by the input to be defined. Converting an inverse cotangent to a reciprocal-input expression therefore needs particular care at zero.

Note: For every real x, tan⁻¹x + cot⁻¹x = π/2. This complementary relation respects both principal ranges and includes the zero-input case.

How are inverse secant and inverse cosecant evaluated?

The original functions satisfy sec θ = 1/cos θ and cosec θ = 1/sin θ wherever these reciprocals are defined. Consequently, inverse secant and inverse cosecant can be evaluated using cosine and sine equations with reciprocal inputs.

First require x ≤ −1 or x ≥ 1. Then 1/x belongs to [−1, 1] and is non-zero. To evaluate sec⁻¹x, find the angle whose cosine is 1/x in the secant principal range. To evaluate cosec⁻¹x, use sine instead.

Which reciprocal conversions preserve the branch?

The principal-value relations are sec⁻¹x = cos⁻¹(1/x) and cosec⁻¹x = sin⁻¹(1/x), for x ≤ −1 or x ≥ 1. Their ranges fit the required exclusions automatically because the reciprocal input cannot be zero.

Worked example 6. Find cosec⁻¹2.

Answer: Put θ = cosec⁻¹2. Then cosec θ = 2, so sin θ = 1/2. The angle π/6 belongs to [−π/2, π/2] − {0}. Hence cosec⁻¹2 = π/6.

Worked example 7. Find sec⁻¹(2/√3).

Answer: The equation sec θ = 2/√3 becomes cos θ = √3/2. The angle π/6 belongs to [0, π] − {π/2}, so sec⁻¹(2/√3) = π/6.

The reciprocal conversion changes the input number, not the inverse notation into a reciprocal of an angle. After the conversion, the output remains an angle. Check its principal range exactly as for sine or cosine inverse, including any excluded angle.

What do the graphs of inverse trigonometric functions show?

In a graph y = f(x), x is the horizontal coordinate and y is the vertical coordinate. If (a, b) is a point of an invertible function, with a and b denoting its two coordinates, then (b, a) is a point of its inverse.

The coordinate exchange is a reflection, or mirror image, in the line y = x. For a trigonometric inverse, first select the appropriate restricted original graph. Reflecting that branch gives the corresponding single-valued inverse graph.

How are the sine and cosine branches represented?

The three panels show the sine curve, its inverse branches, and both curves with the line y = x. The darker inverse portion marks the principal branch from output −π/2 to π/2.

See Fig. 2.1 in your NCERT textbook

The first panel shows the cosine curve. The second shows inverse branches, with the darker principal portion descending from input −1 and output π to input 1 and output 0.

See Fig. 2.2 in your NCERT textbook

The inverse-sine principal graph rises as the input increases, while the inverse-cosine principal graph falls. Both occupy the horizontal input interval [−1, 1], but their vertical output intervals differ. The axes show why a shared domain does not imply a shared range.

What do the remaining four graphs show?

Each figure pairs the original function with its inverse branches. The inverse curves occur outside the central input gap between −1 and 1; darker portions distinguish the principal branches.

See Figs. 2.3 and 2.4 in your NCERT textbook

Each figure shows original and inverse branches with dashed boundary lines. The darker tangent-inverse branch rises through the origin, the point (0, 0). The darker cotangent-inverse branch falls through output π/2 at input zero.

See Figs. 2.5 and 2.6 in your NCERT textbook

A horizontal asymptote is a horizontal line approached by a curve as the input increases or decreases without bound. The tangent-inverse graph approaches output levels −π/2 and π/2; the cotangent-inverse graph approaches 0 and π. These levels are excluded from their respective ranges.

Read a sketch together with its domain and range. Additional inverse branches in a drawing do not become additional outputs of the principal inverse function. The selected darker portion supplies the values used when no alternative branch is specified.

When do a trigonometric function and its inverse cancel?

A composition applies one function to the output of another. In sin(sin⁻¹x), the inverse sine is applied first. In sin⁻¹(sin θ), the sine is applied first. Reversing the order changes which restriction must be checked.

Identity: Composition with the inverse applied first

sin(sin⁻¹x) = x for −1 ≤ x ≤ 1. The inverse selects the principal angle having sine x, and applying sine returns x. Similarly, cos(cos⁻¹x) = x on [−1, 1], and tan(tan⁻¹x) = x for every real x.

The same reasoning gives cot(cot⁻¹x) = x for every real x. It gives sec(sec⁻¹x) = x and cosec(cosec⁻¹x) = x when x ≤ −1 or x ≥ 1. Each statement requires the inner inverse function to exist.

Identity: Composition with the trigonometric function applied first

sin⁻¹(sin θ) = θ requires −π/2 ≤ θ ≤ π/2. Likewise, cos⁻¹(cos θ) = θ requires 0 ≤ θ ≤ π, and tan⁻¹(tan θ) = θ requires −π/2 < θ < π/2.

For cotangent, secant and cosecant, the same cancellation requires θ to belong to the corresponding principal range in the table. Outside that range, seek the principal angle with the same trigonometric value. Do not cancel the symbols before checking the angle.

Worked example 8. Evaluate sin⁻¹(sin(3π/5)).

Answer: The angle 3π/5 is outside [−π/2, π/2]. Since sin(3π/5) = sin(π − 3π/5) = sin(2π/5), and 2π/5 is inside that interval, the principal value is 2π/5.

The calculation changes the angle while preserving its sine. This is why inverse sine does not return every angle originally supplied to sine. It returns the unique angle in its chosen branch with the required sine value.

Note: Inverse-trigonometric identities are valid within the principal value branches and wherever the expressions are defined. Some results may not be valid for all values of the domains. Preserve both the definition conditions and the branch conditions.

How are complementary, negative-input and conversion identities used?

Two angles are complementary when their sum is π/2. Complementary trigonometric relations connect several pairs of inverse functions. Each equality below refers to principal values, with its input restrictions forming part of the statement.

Identity: Complementary principal values

For −1 ≤ x ≤ 1, sin⁻¹x + cos⁻¹x = π/2. For every real x, tan⁻¹x + cot⁻¹x = π/2. For x ≤ −1 or x ≥ 1, sec⁻¹x + cosec⁻¹x = π/2.

To justify the first relation, let α, the Greek letter alpha, denote sin⁻¹x. Then sin α = x. The angle π/2 − α lies in [0, π] and its cosine equals x. It therefore equals cos⁻¹x, establishing the sum.

The angle-range step explains why the proof works for negative as well as positive x. The secant-cosecant relation follows by using reciprocal inputs in the same sine-cosine relation. The tangent-cotangent relation follows by applying the corresponding complementary-angle relation on their principal ranges.

How does changing the input sign affect the output?

Within their domains, sin⁻¹(−x) = −sin⁻¹x, tan⁻¹(−x) = −tan⁻¹x, and cosec⁻¹(−x) = −cosec⁻¹x. Their principal ranges are symmetric about zero, so changing the angle's sign preserves the required branch.

The other three follow different rules: cos⁻¹(−x) = π − cos⁻¹x, cot⁻¹(−x) = π − cot⁻¹x, and sec⁻¹(−x) = π − sec⁻¹x, for inputs in their respective domains. These outputs remain in the relevant non-negative principal ranges.

Which conversions need extra restrictions?

For −1 < x < 1, sin⁻¹x = tan⁻¹(x/√(1 − x²)). Here x² means x multiplied by itself. The denominator is positive, and the angle belongs to both required principal ranges.

For 0 ≤ x ≤ 1, sin⁻¹x = cos⁻¹√(1 − x²). The non-negative restriction matters because the cosine inverse on the right returns a non-negative angle. The combined sine-cosine-tangent conversion is therefore valid for 0 ≤ x < 1.

The reciprocal identity sin⁻¹x = cosec⁻¹(1/x) requires −1 ≤ x ≤ 1 and x ≠ 0, where ≠ means “not equal to”. Similarly, cos⁻¹x = sec⁻¹(1/x) requires the same input conditions. A conversion involving 1/x cannot include x = 0.

How are sums and differences of inverse functions simplified?

For sum and difference identities, define α and β, the Greek letters alpha and beta, as the two principal angles. First calculate the trigonometric value of their sum or difference. Then check whether that combined angle belongs to the range of the inverse on the proposed right-hand side.

How do the sine and cosine formulae work?

Let α = sin⁻¹x and β = sin⁻¹y, where x and y are real inputs in [−1, 1]. Then cos α = √(1 − x²) and cos β = √(1 − y²), because cosine is non-negative on the inverse-sine principal range.

The addition formula gives sin(α + β) = x√(1 − y²) + y√(1 − x²). Thus α + β = sin⁻¹(x√(1 − y²) + y√(1 − x²)) provided −π/2 ≤ α + β ≤ π/2.

For the difference, sin(α − β) = x√(1 − y²) − y√(1 − x²). Consequently α − β equals the inverse sine of this expression provided −π/2 ≤ α − β ≤ π/2. The sign changes, but the branch-check principle stays the same.

Now let α = cos⁻¹x and β = cos⁻¹y. The sine values are the non-negative roots √(1 − x²) and √(1 − y²). Hence cos(α + β) = xy − √(1 − x²)√(1 − y²).

Taking the principal cosine inverse returns α + β when 0 ≤ α + β ≤ π. For a difference, cos(α − β) = xy + √(1 − x²)√(1 − y²), and cancellation requires 0 ≤ α − β ≤ π.

Identity: Tangent inverse addition and subtraction

For real x and y with xy < 1, tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)). Here xy denotes multiplication. The condition keeps the angle sum inside the tangent-inverse principal range and makes the denominator non-zero.

For real x and y with xy > −1, tan⁻¹x − tan⁻¹y = tan⁻¹((x − y)/(1 + xy)). These strict conditions exclude the zero-denominator boundaries.

To see the range condition for addition, put α = tan⁻¹x and β = tan⁻¹y. Both cos α and cos β are positive. The formula cos(α + β) = cos α cos β(1 − xy) makes the cosine of the sum positive when xy < 1.

Since the sum lies between −π and π, that positive cosine places it between −π/2 and π/2. The subtraction argument uses cos(α − β) = cos α cos β(1 + xy). The denominator condition therefore controls the branch as well as the arithmetic.

How are double-angle and triple-angle expressions handled?

In expressions such as sin²θ, cos²θ and tan²θ, the superscript 2 squares the trigonometric value; similarly, tan³θ cubes that value. A double-angle formula relates a function of twice an angle to functions of the original angle. A triple-angle formula does the same for three times an angle. In inverse-trigonometric work, the final angle must still be checked against the selected inverse range.

What are the sine and cosine doubling results?

For −1/√2 ≤ x ≤ 1/√2, 2sin⁻¹x = sin⁻¹(2x√(1 − x²)). Put θ = sin⁻¹x. Its cosine is √(1 − x²), so sin 2θ = 2x√(1 − x²), and the stated restriction places 2θ in [−π/2, π/2].

For 1/√2 ≤ x ≤ 1, 2cos⁻¹x = sin⁻¹(2x√(1 − x²)). Here θ = cos⁻¹x lies between 0 and π/4. Its double lies between 0 and π/2, so inverse sine returns that double angle.

Another useful form is 2cos⁻¹x = cos⁻¹(2x² − 1) for 0 ≤ x ≤ 1. This follows from cos 2θ = 2cos²θ − 1, with 2θ in [0, π]. The different right-hand inverse explains the different permissible input interval.

Worked example 9. Prove that 2sin⁻¹(3/5) = tan⁻¹(24/7).

Answer: Put θ = sin⁻¹(3/5). Then cos θ = 4/5 and tan θ = 3/4. Also 0 < θ < π/4, so 0 < 2θ < π/2. Therefore tan 2θ = 2(3/4)/(1 − 9/16) = 24/7, and the branch check gives 2θ = tan⁻¹(24/7).

How are tangent doubles and triples expressed?

For −1 < x < 1, 2tan⁻¹x = tan⁻¹(2x/(1 − x²)). Put θ = tan⁻¹x and apply tan 2θ = 2tan θ/(1 − tan²θ). The input interval ensures −π/2 < 2θ < π/2.

For −1/√3 < x < 1/√3, 3tan⁻¹x = tan⁻¹((3x − x³)/(1 − 3x²)). The symbol x³ means x multiplied by itself three times. This interval places θ = tan⁻¹x between −π/6 and π/6, so its triple is in the principal tangent range.

The triple-angle expression comes from tan 3θ = (3tan θ − tan³θ)/(1 − 3tan²θ). Outside the stated intervals, the tangent of the combined angle can still be calculated where defined, but taking its principal inverse need not return the original double or triple.

How can an inverse-trigonometric equation be solved and checked?

An equation asks for input values making two expressions equal. With inverse functions, first establish every input domain. Next establish the output intervals, apply a suitable trigonometric identity, and finally substitute each candidate into the original equation.

An extraneous candidate is a value admitted by a transformed equation that fails the original equation. Applying a trigonometric function can lose angle information because that function repeats values. Substitution into the original expression restores the necessary domain and branch checks.

What does a complete solution look like?

Worked example 10. Solve tan⁻¹((1 − x)/(1 + x)) = (1/2)tan⁻¹x, given x > 0.

Answer: Put θ = tan⁻¹x, so 0 < θ < π/2. The tangent subtraction formula gives (1 − x)/(1 + x) = tan(π/4 − θ). Since −π/4 < π/4 − θ < π/4, the left-hand inverse equals π/4 − θ.

The equation becomes π/4 − θ = θ/2. Therefore θ = π/6 and x = tan(π/6) = 1/√3. For this value, the left side equals π/4 − π/6 = π/12, while the right side equals (1/2)(π/6) = π/12.

The restriction x > 0 performs two jobs: it excludes a zero denominator at x = −1 and fixes the interval for θ. The subtraction angle then lies inside the tangent-inverse range, so replacing the inverse expression by that angle is justified.

1. Domain: Record the admissible inputs, including square-root and denominator restrictions introduced during the calculation.
2. Angle: Give each inverse expression an angle name and retain its principal range.
3. Identity: Apply the relevant trigonometric formula, checking the combined angle before inverse cancellation.
4. Verification: Substitute the candidate into the original equation and compare both principal-value expressions.

For an identity proof, the same sequence works with a variable instead of a numerical candidate. State the input interval, establish the angle interval, perform the algebra, and connect the resulting angle to the correct principal inverse.

Glossary

• Domain — The set of input values for which the specified function is defined.
• Range — The set of output values actually obtained from the permitted inputs.
• Codomain — The specified target set containing all possible outputs of a function.
• One-one function — A function in which distinct inputs produce distinct output values.
• Onto function — A function whose range is the whole of its specified codomain.
• Bijective function — A function that is both one-one and onto and therefore has an inverse.
• Inverse function — A function reversing the input-output correspondence of an original bijective function.
• Branch — An inverse function obtained by choosing a suitable restricted domain of the original function.
• Principal value branch — The conventional inverse branch used when no alternative branch is specified.
• Principal value — The angle returned within the range of the chosen principal inverse branch.
• Composition — The application of one function to the output produced by another function.
• Reciprocal — The value obtained by dividing one by a specified non-zero quantity.
• Complementary angles — Two angles whose sum is π/2 when measured in radians.
• Horizontal asymptote — A horizontal line approached by a graph as its input increases or decreases without bound.
• Extraneous candidate — A value satisfying a transformed equation but failing the original equation or its restrictions.

Common errors and misconceptions

• Misconception: sin⁻¹x means 1/sin x. Correct: sin⁻¹x returns a principal angle; the reciprocal is written (sin x)⁻¹. These are different operations with different input and output meanings.
• Misconception: Any angle with the required trigonometric value is an inverse-function answer. Correct: The angle must belong to the specified branch. Without another branch being named, use the principal value branch.
• Misconception: Inverse tangent and inverse cotangent have the same principal range. Correct: Their ranges are (−π/2, π/2) and (0, π), respectively. A negative input therefore produces different angle-sign behaviour.
• Misconception: sin⁻¹(sin θ) equals θ for every angle. Correct: Direct cancellation requires θ in [−π/2, π/2]. Otherwise find the principal angle that has the same sine.
• Misconception: The range of inverse tangent includes ±π/2, where ± means either sign. Correct: Both endpoints are excluded. Inverse sine includes these endpoints, so the bracket types must be distinguished.
• Misconception: sin⁻¹x = cos⁻¹√(1 − x²) for every x in [−1, 1]. Correct: This form requires 0 ≤ x ≤ 1; negative inputs would give incompatible principal-angle signs.
• Misconception: Inverse-trigonometric sum formulae hold without conditions. Correct: Check the input domains, denominator and combined-angle range. A trigonometric addition formula alone does not justify principal inverse cancellation.
• Misconception: Solving a transformed equation completes the calculation. Correct: Check each candidate in the original equation, using principal values and every stated input restriction.

Exam-style questions with model answers

Q1. State the domain and principal value range of sin⁻¹x and tan⁻¹x. [2 marks]

1. The domain of sin⁻¹x is [−1, 1], and its principal value range is [−π/2, π/2], including both endpoints.
2. The domain of tan⁻¹x is all real numbers, R, and its principal value range is (−π/2, π/2), excluding both endpoints.

Q2. Find the principal value of cot⁻¹(−1/√3), showing the range check. [3 marks]

1. Let θ = cot⁻¹(−1/√3). Then cot θ = −1/√3, and the principal cotangent-inverse range requires 0 < θ < π.
2. Since cot(π/3) = 1/√3, the supplementary angle 2π/3 has cotangent −1/√3. Supplementary angles are angles whose sum is π.
3. The angle 2π/3 lies in the required interval (0, π). Therefore cot⁻¹(−1/√3) = 2π/3; a negative angle would not satisfy this principal-range requirement.

Q3. Evaluate tan⁻¹1 + cos⁻¹(−1/2) + sin⁻¹(−1/2), using principal values. [3 marks]

1. The principal value tan⁻¹1 is π/4, because tan(π/4) = 1 and π/4 belongs to the open interval (−π/2, π/2).
2. The principal cosine and sine values are cos⁻¹(−1/2) = 2π/3 and sin⁻¹(−1/2) = −π/6. These belong to [0, π] and [−π/2, π/2], respectively.
3. Adding the selected angles gives π/4 + 2π/3 − π/6 = (3π + 8π − 2π)/12 = 3π/4. Each inverse has been evaluated before the addition.

Q4. Evaluate sin⁻¹(sin(3π/5)) and explain why direct cancellation fails. [4 marks]

1. The principal inverse-sine range is [−π/2, π/2]. The supplied angle 3π/5 is greater than π/2, so it cannot itself be the principal answer.
2. Use the identity sin(π − θ) = sin θ, where θ denotes an angle. It gives sin(3π/5) = sin(2π/5).
3. The replacement angle 2π/5 lies between 0 and π/2. It therefore belongs to the principal inverse-sine range.
4. Consequently sin⁻¹(sin(3π/5)) = sin⁻¹(sin(2π/5)) = 2π/5. Direct cancellation fails because the original angle is outside the required branch.

Q5. Prove that 2sin⁻¹(3/5) = tan⁻¹(24/7), using principal values and justifying the angle range. [5 marks]

1. Let θ = sin⁻¹(3/5). Then sin θ = 3/5 and 0 < θ < π/2, as the input is positive and less than one.
2. Since cosine is positive in this interval, cos θ = √(1 − 9/25) = 4/5. Therefore tan θ = (3/5)/(4/5) = 3/4.
3. Because tan θ = 3/4 < 1 on this increasing tangent branch, θ < π/4. Hence the double angle satisfies 0 < 2θ < π/2.
4. Apply the double-angle identity: tan 2θ = 2tan θ/(1 − tan²θ) = (3/2)/(1 − 9/16) = (3/2)/(7/16) = 24/7.
5. The double angle is inside the tangent-inverse principal range. Thus 2θ = tan⁻¹(24/7), giving the required equality after replacing θ by sin⁻¹(3/5).

Q6. For a real number x with −1/√2 ≤ x ≤ 1/√2, prove sin⁻¹(2x√(1 − x²)) = 2sin⁻¹x. [5 marks]

1. Set θ = sin⁻¹x, so x = sin θ. The given input interval places θ between −π/4 and π/4 on the principal inverse-sine branch.
2. Cosine is non-negative throughout this angle interval. Therefore √(1 − x²) = √(1 − sin²θ) = cos θ, with the non-negative square root correctly selected.
3. Substitute these expressions into the inner argument: 2x√(1 − x²) = 2sin θ cos θ = sin 2θ, using the sine double-angle formula.
4. Doubling the established angle bounds gives −π/2 ≤ 2θ ≤ π/2. Thus the double angle lies within the closed principal range of inverse sine.
5. It follows that sin⁻¹(2x√(1 − x²)) = sin⁻¹(sin 2θ) = 2θ = 2sin⁻¹x. The supplied interval is what permits this final cancellation.

Q7. Solve tan⁻¹((1 − x)/(1 + x)) = (1/2)tan⁻¹x for the real number x, given x > 0. Check your result. [6 marks]

1. Let θ = tan⁻¹x. Since x > 0, the principal angle satisfies 0 < θ < π/2, and tan θ = x.
2. The tangent subtraction identity gives tan(π/4 − θ) = (1 − tan θ)/(1 + tan θ) = (1 − x)/(1 + x). The denominator is positive.
3. The angle π/4 − θ lies in (−π/4, π/4), entirely inside the tangent-inverse principal range. Therefore the left side of the given equation equals π/4 − θ.
4. The equation becomes π/4 − θ = θ/2. Rearranging gives 3θ/2 = π/4 and hence θ = π/6.
5. Thus x = tan θ = tan(π/6) = 1/√3. This value is positive and therefore meets the input restriction in the question.
6. Substitution gives π/4 − π/6 = π/12 on the left and (1/2)(π/6) = π/12 on the right. The checked solution is x = 1/√3.

Q8. Explain how the principal graph of y = sin⁻¹x is obtained from the sine graph. State its domain and range. [4 marks]

1. Restrict the original sine function to the angle interval [−π/2, π/2]. On this interval it is one-one and maps onto [−1, 1].
2. Interchange the coordinates of each point on that restricted graph. Geometrically, this reflects the selected sine branch in the line y = x.
3. The original output interval becomes the inverse input interval. Therefore the domain of the principal inverse-sine graph is [−1, 1].
4. The original restricted input interval becomes the inverse output interval. Its range is [−π/2, π/2], and this reflected branch is the principal graph.

Key takeaways

• Restricting a trigonometric function to a suitable one-one domain permits an inverse that returns a uniquely selected angle.
• The principal value branch supplies the inverse angle whenever the question does not specify an alternative branch.
• Memorise each domain together with its principal range, including closed endpoints, open endpoints and excluded individual angles.
• Inverse sine and inverse cosine accept the same inputs but return angles in different principal ranges.
• Inverse tangent and inverse cotangent accept every real input, but their principal ranges require different choices for negative inputs.
• Inverse notation is different from reciprocal notation: sin⁻¹x returns an angle, whereas (sin x)⁻¹ means one divided by sine.
• A trigonometric addition formula determines a trigonometric value; the combined-angle range determines whether principal inverse cancellation is justified.
• Check denominator restrictions, square-root signs and principal ranges when simplifying expressions, proving identities or solving inverse-trigonometric equations.

Test yourself

Why must the sine domain be restricted before defining its inverse?

Sine repeats values on its full real domain. Restriction to [−π/2, π/2] makes it one-one onto [−1, 1], allowing a unique inverse.

What is the principal range of cot⁻¹x?

It is (0, π), excluding both endpoints. This differs from the principal range of inverse tangent.

What is the principal value of cos⁻¹(−1/2)?

It is 2π/3, because its cosine is −1/2 and it belongs to the interval [0, π].

What does (sin x)⁻¹ mean?

It means the reciprocal 1/sin x, defined when sin x is non-zero. It does not mean inverse sine.

What is sin⁻¹(sin(3π/5))?

It is 2π/5, the angle in [−π/2, π/2] having the same sine as 3π/5.

For which inputs can sec⁻¹x be evaluated as a real principal angle?

The input must satisfy x ≤ −1 or x ≥ 1. Values strictly between −1 and 1 are excluded.

What condition permits tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy))?

For real x and y, require xy < 1. This places the sum inside the principal tangent-inverse range.

Why is a range check needed after applying a double-angle formula?

The doubled angle can leave the principal inverse range. Taking an inverse then returns a different principal angle with the same trigonometric value.