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Inverse Trigonometric Functions · Lesson 4 of 4

Chapter Summary and Practice

“Connect all six principal inverses, range checks and simplification methods through mixed reasoning and applications.”

Learning Objectives

• Connect bijective restrictions, principal values, graph reflection and all six domain-and-range choices. • Select principal representatives for out-of-range angles. • Prove inverse identities using ratios together with an explicit angle-range check. • Solve inverse-trigonometric equations while preserving domains and avoiding extra solutions. • Apply the chapter’s methods to mixed expressions and interpret the historical height-and-distance connection.

One chapter built around a single choice

The chapter began with a practical difficulty: many angles can have the same trigonometric ratio. Restricting the angle interval made each function reversible, and choosing a principal branch gave one agreed output. Every later property depends on that choice, so a useful review should connect the range table, graphs, identities and equation methods rather than memorise them separately.

Think of an inverse-trigonometric expression as an instruction with two parts: match a ratio, then return its angle in a specified interval. A correct ratio alone is not enough. Likewise, a plausible angle in the right interval is not enough unless its sine, cosine or other ratio matches the input.

Chapter ideaConnection to the rest of the chapterQuestion to ask
Bijective restrictionMakes reversing input-output pairs possible.Does each allowed ratio occur exactly once on this chosen angle set?
Principal branch and valueSelects one conventional output from repeating angle possibilities.Which angle interval is this inverse allowed to return?
Domains, exclusions and endpointsCome from the original ratio range and the selected angle domain.Is the input allowed, and is an endpoint or removed angle being used?
Graph reflectionExchanges (a, b) with (b, a) across y = x.Have I restricted the original function before reflecting it?
Inverse compositionEither recovers an allowed ratio or selects a principal angle.Am I starting from a ratio or from an angle?
Angle identities and substitutionsTurn algebraic expressions into ordinary trigonometric ratios.Are roots signed correctly, denominators nonzero, and the final angle in range?
Mixed proofs and equationsCombine equality of ratios with domain and angle constraints.Have all proposed solutions been checked in the original expression?
ApplicationsUse a known ratio to determine an angle.What physical or geometric interval selects the angle?

The six principal functions as a connected family

The table below is more useful when read in pairs. Sine and cosine inverses share an input domain; cosecant and secant inverses share a different input domain; tangent and cotangent inverses accept every real input. Within each pair, the output intervals and graph behaviour differ.

Principal inverseInput domainOutput range
sin⁻¹ x[−1, 1][−π/2, π/2]
cos⁻¹ x[−1, 1][0, π]
cosec⁻¹ x(−∞, −1] ∪ [1, ∞)[−π/2, π/2] excluding 0
sec⁻¹ x(−∞, −1] ∪ [1, ∞)[0, π] excluding π/2
tan⁻¹ xℝ(−π/2, π/2)
cot⁻¹ xℝ(0, π)

Inverse sine and inverse tangent increase and preserve the sign of an input. Principal inverse cosecant also preserves sign, though its two graph pieces decrease separately. Inverse cosine and inverse cotangent decrease and return nonnegative or positive angles. Inverse secant returns a first-quadrant angle for a positive input and a second-quadrant angle for a negative input; its two pieces increase separately.

A horizontal or vertical asymptote signals an approached value rather than an included endpoint. Inverse tangent approaches ±π/2, inverse cotangent approaches 0 and π, inverse cosecant approaches 0, and inverse secant approaches π/2. The sine and cosine inverse endpoints are included. In the reciprocal inverses, input values ±1 are included, even though the input interval between them is excluded.

Common mistake

Do not confuse an inverse with a reciprocal or confuse an inverse input with its output. Sin⁻¹ 0 is an angle, 0, while 1/sin 0 is undefined. Also, inverse cotangent of a negative input must still lie in (0, π); it does not follow inverse tangent’s negative-angle convention.

Bring the angle back into the principal interval

When an ordinary trigonometric function comes first, its ratio no longer records the original turn or branch. The inverse returns the angle representing that ratio in its own principal interval. Periodicity and symmetry let us find that representative exactly.

Example — A supplementary sine angle

Problem
Evaluate sin⁻¹(sin(3π/5)).

  1. 1.The inner angle 3π/5 is larger than π/2, so it is outside the principal inverse-sine range.
  2. 2.Sine has the same value at supplementary angles: sin(3π/5) = sin(π − 3π/5) = sin(2π/5).
  3. 3.The replacement angle 2π/5 is in [−π/2, π/2].
  4. 4.Therefore sin⁻¹(sin(3π/5)) = 2π/5. The original angle was replaced, not recovered.
Two angles, the same sine3π/52π/5xyπ − 3π/5 = 2π/5 lies in the principal inverse-sine range.
Supplementary angles and a principal representative— The two points on the unit circle have the same vertical coordinate, hence the same sine. Only 2π/5 lies in the principal sine interval.
Example — Period reduction followed by range selection

Problem
Evaluate cos⁻¹(cos(13π/6)) and tan⁻¹(tan(7π/6)).

  1. 1.Cosine has period 2π. Subtract 2π from 13π/6 to obtain π/6, which lies in [0, π]. Thus cos⁻¹(cos(13π/6)) = π/6.
  2. 2.Tangent has period π. Subtract π from 7π/6 to obtain π/6, which lies in (−π/2, π/2). Thus tan⁻¹(tan(7π/6)) = π/6.
  3. 3.The periods differ. Using π as a cosine period would change its sign rather than preserve its value.
  4. 4.After reducing by a period, always check the resulting interval. Period reduction alone does not guarantee a principal representative for every starting angle.

Prove angle equalities by checking a ratio and a range

If two angles have the same tangent, they may differ by a multiple of π. If two have the same sine, they may be supplementary or belong to different turns. An inverse identity is proved only when the proposed angle has the required ratio and belongs to an interval where that ratio is one-to-one.

For positive inputs in inverse sine or inverse cosine, right-triangle ratios give exact missing sides through the Pythagorean relationship. When adding two such acute angles, the sum is between 0 and π. Computing a positive cosine then places that sum in (0, π/2), which is the range needed for an inverse tangent or a positive inverse sine.

Example — A double-angle proof with a range check

Problem
Prove 2sin⁻¹(3/5) = tan⁻¹(24/7).

  1. 1.Let θ = sin⁻¹(3/5). It is acute, and cos θ = √(1 − 9/25) = 4/5.
  2. 2.Compute sin(2θ) = 2(3/5)(4/5) = 24/25 and cos(2θ) = (4/5)² − (3/5)² = 7/25.
  3. 3.Since 0 < 2θ < π and cos(2θ) > 0, the doubled angle lies in (0, π/2).
  4. 4.Its tangent is (24/25)/(7/25) = 24/7.
  5. 5.This ratio together with 2θ in the principal tangent range proves that tan⁻¹(24/7) = 2θ = 2 sin⁻¹(3/5).
Example — Adding two inverse-sine angles

Problem
Prove sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36).

  1. 1.Let α = sin⁻¹(8/17) and β = sin⁻¹(3/5). Both are acute. Their cosines are 15/17 and 4/5 respectively.
  2. 2.Use sine addition: sin(α + β) = (8/17)(4/5) + (15/17)(3/5) = 77/85.
  3. 3.Use cosine addition: cos(α + β) = (15/17)(4/5) − (8/17)(3/5) = 36/85.
  4. 4.The sum is in (0, π), and its cosine is positive, so 0 < α + β < π/2.
  5. 5.Its tangent is 77/36. The sum is in the principal inverse-tangent range, so it equals tan⁻¹(77/36).

The next three combinations use the same method. In each row both component angles are acute, so their sum is initially known to lie in (0, π). The positive cosine listed in the table then proves that the sum is acute. This is the essential range check when converting the computed sine into an inverse sine or the computed tangent into an inverse tangent.

Angle sum Asin Acos APrincipal inverse result
cos⁻¹(4/5) + cos⁻¹(12/13)56/6533/65A = cos⁻¹(33/65)
cos⁻¹(12/13) + sin⁻¹(3/5)56/6533/65A = sin⁻¹(56/65)
sin⁻¹(5/13) + cos⁻¹(3/5)63/6516/65A = tan⁻¹(63/16)

For the first row, the missing positive sines are 3/5 and 5/13. Hence cos A = (4/5)(12/13) − (3/5)(5/13) = 33/65, and sin A = (3/5)(12/13) + (4/5)(5/13) = 56/65. For the second row, the two sines are 5/13 and 3/5, with cosines 12/13 and 4/5; addition gives the same pair. For the third row, the sine and cosine pairs are (5/13, 12/13) and (4/5, 3/5), giving sin A = (15 + 48)/65 and cos A = (36 − 20)/65. The ratio values now have a fully justified angle interpretation.

Half-angle substitutions link radicals to principal angles

Radical identities in the mixed problems use the same sign controls as the earlier lesson. Put the input in terms of a principal angle and take square roots only after locating that angle. This turns the radicals into positive sine and cosine combinations, revealing a tangent or cotangent.

Example — A reciprocal double-angle relationship

Problem
Prove tan⁻¹(√x) = ½cos⁻¹((1 − x)/(1 + x)) for 0 ≤ x ≤ 1.

  1. 1.Let θ = tan⁻¹√x. The input range gives 0 ≤ θ ≤ π/4, so 0 ≤ 2θ ≤ π/2.
  2. 2.The cosine double-angle identity in tangent form is cos(2θ) = (1 − tan² θ)/(1 + tan² θ). Substituting tan² θ = x gives (1 − x)/(1 + x).
  3. 3.The denominator is positive, and 2θ lies inside [0, π], so cos⁻¹((1 − x)/(1 + x)) = 2θ.
  4. 4.Dividing by 2 gives the required identity, including its endpoints x = 0 and x = 1.
Example — A sum and difference of square roots

Problem
Prove cot⁻¹((√(1 + sin x) + √(1 − sin x))/(√(1 + sin x) − √(1 − sin x))) = x/2 for 0 < x < π/4.

  1. 1.Let θ = x/2, so 0 < θ < π/8. Write c = cos θ and s = sin θ; here c > s > 0.
  2. 2.Since 1 + sin x = (c + s)² and 1 − sin x = (c − s)², their nonnegative roots are c + s and c − s.
  3. 3.The quotient becomes [(c + s) + (c − s)]/[(c + s) − (c − s)] = 2c/(2s) = cot θ. The denominator is positive.
  4. 4.The angle θ lies in the principal cotangent range (0, π), so the expression equals θ = x/2.
Example — A radical difference revealing a shifted half angle

Problem
Prove tan⁻¹((√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))) = π/4 − ½cos⁻¹x for −1/√2 ≤ x ≤ 1.

  1. 1.Put θ = (1/2) cos⁻¹ x. The input range gives 0 ≤ θ ≤ 3π/8, and x = cos(2θ).
  2. 2.Both sin θ and cos θ are nonnegative. Thus √(1 + x) = √2 cos θ and √(1 − x) = √2 sin θ.
  3. 3.The quotient becomes (cos θ − sin θ)/(cos θ + sin θ) = tan(π/4 − θ). Its denominator is positive throughout this interval.
  4. 4.The shifted angle π/4 − θ lies in [−π/8, π/4], inside (−π/2, π/2).
  5. 5.The outer inverse tangent therefore returns π/4 − θ = π/4 − (1/2) cos⁻¹ x.

Solve equations without losing the original restrictions

Taking an ordinary trigonometric function of both sides can turn an inverse equation into an algebraic or trigonometric equation. This is useful, but it may discard information about the permitted angle interval. Record the original domain first, and either keep both sides in a one-to-one interval or verify every proposed solution afterwards.

Example — An equation that simplifies after taking tangent

Problem
Solve 2tan⁻¹(cos x) = tan⁻¹(2cosec x) for real x.

  1. 1.The original equation requires sin x ≠ 0. Therefore |cos x| < 1, so 2 tan⁻¹(cos x) lies strictly between −π/2 and π/2. The right side lies in that interval too.
  2. 2.Tangent is one-to-one there. Applying it gives 2 cos x/(1 − cos² x) = 2/sin x; all denominators are nonzero under the original restriction.
  3. 3.Use 1 − cos² x = sin² x and multiply by sin² x/2 to obtain cos x = sin x.
  4. 4.The solutions are x = π/4 + kπ, where k is any integer. Each has nonzero sine and satisfies cos x = sin x.
  5. 5.For these values, both original angle expressions belong to the same tangent principal interval and have equal tangents. They are therefore equal, so every listed solution is valid.
Example — Reducing an inverse equation to a linear angle equation

Problem
Solve tan⁻¹((1 − x)/(1 + x)) = ½tan⁻¹x with x > 0.

  1. 1.Let θ = tan⁻¹ x. Since x > 0, 0 < θ < π/2. The tangent subtraction formula gives (1 − x)/(1 + x) = tan(π/4 − θ).
  2. 2.The angle π/4 − θ is in (−π/4, π/4), so its inverse tangent is exactly π/4 − θ.
  3. 3.The equation becomes π/4 − θ = θ/2. Hence (3/2)θ = π/4 and θ = π/6.
  4. 4.Thus x = tan(π/6) = 1/√3, which is positive.
  5. 5.At this value, the original left side is tan⁻¹(2 − √3) = π/12, equal to half of π/6 on the right. The solution passes the original equation.
Example — Use bounds before trying algebra

Problem
Solve sin⁻¹(1 − x) − 2sin⁻¹x = π/2 over the reals.

  1. 1.The two inverse-sine domains require −1 ≤ x ≤ 1 and −1 ≤ 1 − x ≤ 1. Together they give 0 ≤ x ≤ 1.
  2. 2.Rewrite the equation as sin⁻¹(1 − x) = π/2 + 2 sin⁻¹ x.
  3. 3.On the allowed x interval, the left side is at most π/2, while the right side is at least π/2. Equality requires sin⁻¹ x = 0.
  4. 4.Therefore x = 0. Substitution gives sin⁻¹ 1 − 2 sin⁻¹ 0 = π/2, so it is valid.
  5. 5.This bound argument avoids squaring or applying sine to an unrestricted equation, either of which could produce unnecessary candidate solutions.
Example — A sine of an inverse tangent with either sign

Problem
Find sin(tan⁻¹x) for |x| < 1.

  1. 1.Let θ = tan⁻¹ x. Its principal range gives cos θ > 0, whether x is positive, negative or zero.
  2. 2.Use 1 + tan² θ = sec² θ to obtain cos θ = 1/√(1 + x²).
  3. 3.Then sin θ = tan θ cos θ = x/√(1 + x²).
  4. 4.The answer has the same sign as x, as it should. The reasoning actually works for every real x, although the stated problem uses |x| < 1.

Historical context and an angle from a measured ratio

Trigonometry developed through astronomy, geometry and practical measurement. Indian mathematicians including Aryabhata, Brahmagupta, Bhaskara I and Bhaskara II contributed to its development, and mathematical knowledge travelled through Arabic scholarship to Europe. Greek work on geometric measurement provides another part of this history.

The siddhantas, Sanskrit astronomical works, contributed to the development of sine-based methods. Bhaskara I treated sine values beyond 90°, and Bhaskara II gave exact values involving angles such as 18°, 36°, 54° and 72°. The Malayalam work Yuktibhasa contains a proof of the sine addition relationship. The inverse notation sin⁻¹ x and cos⁻¹ x is associated with John Herschel in 1813. These historical connections show how exact ratios, angle identities and notation grew together; they are context rather than additional formulas to memorise here.

Height-and-distance reasoning associated with Thales compares similar triangles. A vertical object and its shadow form a right triangle; another vertical object under the same sunlight gives a similar one. The shared acute angle is the solar altitude, meaning the Sun’s angle above the horizontal. Similar-triangle methods also connect to estimating the distance of a ship at sea and appear in ancient Indian mathematical work.

Similar shadow trianglesLaTeX
H and h are positive vertical heights, S and s are their positive shadow lengths, and α is the common solar altitude in (0, π/2). The inverse tangent converts the measured height-to-shadow ratio into that acute angle.
Similar shadow triangles share one solar angleHSαhsαH/S = h/s = tan αLengths are schematic; both sloping rays have the same direction.
The same angle in two shadow triangles— The two triangles share the direction of the sunlight. Their equal height-to-shadow ratios determine the same acute angle.
Example — Recovering an angle from a shadow

Problem
A vertical pole is 2 m high and casts a horizontal shadow of length 2√3 m. Find the solar altitude.

  1. 1.Use tan α = height/shadow, so tan α = 2/(2√3) = 1/√3.
  2. 2.A solar altitude in this situation is an acute angle, so the required range is (0, π/2).
  3. 3.The principal inverse tangent of 1/√3 is π/6.
  4. 4.Thus α = π/6 radians, or 30°. The measured ratio and the physical angle interval agree.

Check your understanding

These questions connect the whole chapter. Use the principal range for evaluations, a sign check for roots, a ratio-and-range argument for proofs, and the original domain when solving an equation.

Quiz

Quick check

Which fact makes a restricted trigonometric function invertible onto its natural range?

Quick check

Evaluate sin⁻¹(sin(3π/5)).

Quick check

Evaluate cos⁻¹(cos(13π/6)).

Quick check

Why is computing tan A = 77/36 alone insufficient to prove A = tan⁻¹(77/36)?

Quick check

For real x, what is sin(tan⁻¹ x)?

Quick check

For 0 ≤ x ≤ 1, (1/2) cos⁻¹((1 − x)/(1 + x)) equals:

Quick check

Solve sin⁻¹(1 − x) − 2 sin⁻¹ x = π/2 over the reals.

Quick check

Which is the solution of tan⁻¹((1 − x)/(1 + x)) = (1/2) tan⁻¹ x with x > 0?

Quick check

A pole has height h > 0 and shadow length s > 0. Which returns its acute solar-altitude angle?

Quick check

Which pair of endpoint facts is correct?

Practice Problems

Practice Problems
  1. Evaluate cos⁻¹(cos(13π/6)) and tan⁻¹(tan(7π/6)), stating the period and principal interval used.
  2. Prove 2 sin⁻¹(3/5) = tan⁻¹(24/7) by computing both sine and cosine of the doubled angle.
  3. Prove sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36), including the range of the sum.
  4. Prove cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65). Explain why matching cosine is sufficient after the range check.
  5. Prove cos⁻¹(12/13) + sin⁻¹(3/5) = sin⁻¹(56/65), checking that the sum is at most π/2.
  6. Prove tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5) using exact side ratios.
  7. Prove tan⁻¹√x = (1/2) cos⁻¹((1 − x)/(1 + x)) for 0 ≤ x ≤ 1, including both endpoints.
  8. For 0 < x < π/4, prove that the inverse cotangent of (√(1 + sin x) + √(1 − sin x))/(√(1 + sin x) − √(1 − sin x)) equals x/2.
  9. For −1/√2 ≤ x ≤ 1, simplify tan⁻¹((√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))) using x = cos(2θ).
  10. Solve 2 tan⁻¹(cos x) = tan⁻¹(2 cosec x) over the reals. State the excluded inputs before transforming the equation.
  11. Solve tan⁻¹((1 − x)/(1 + x)) = (1/2) tan⁻¹ x for x > 0, and check the proposed value in the original equation.
  12. Find sin(tan⁻¹ x) for |x| < 1, explaining why the square-root denominator has a positive sign.
  13. Solve sin⁻¹(1 − x) − 2 sin⁻¹ x = π/2 using domain and range bounds.
  14. A 3 m pole has a 3 m shadow. Find the solar altitude and explain why a principal inverse tangent gives the relevant angle.
  15. From memory, reconstruct the six-function principal domain-and-range table. Add the original restricted angle set for each function and identify its graph’s endpoints or asymptotes.

Key Takeaways

Key Takeaways

• All six inverse functions begin with a bijective restriction, and a principal branch selects one conventional angle output. • The domain-and-range table encodes allowed ratios, permitted angles, included endpoints and excluded values. • Reflection across y = x exchanges the domain and range of a restricted function. • To simplify or prove an inverse identity, establish the ordinary ratio and then the correct principal angle range. • Square-root signs and nonzero denominators are part of the argument, not optional details. • Periodicity and symmetry replace out-of-range inner angles with principal representatives. • Equation transformations must preserve the original domain and either preserve a one-to-one angle interval or be checked afterwards. • Measured ratios can determine angles when the geometry supplies an appropriate angle interval.

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