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Lesson 3 of 4

Inverse Trigonometric Functions · Lesson 3 of 4

Properties and Simplification

“Use substitutions and angle identities to simplify inverse-trigonometric expressions while keeping every range condition visible.”

Learning Objectives

• Distinguish the two orders of inverse composition and state their restrictions. • Prove double- and triple-angle inverse identities by checking the resulting angle interval. • Use half-angle identities and rationalisation to simplify radicals and ratios. • Choose substitutions whose signs justify square-root simplifications. • Evaluate nested expressions and combinations using principal angles and ordinary trigonometric identities.

A ratio identity is only half an inverse identity

Ordinary trigonometric identities describe equal ratios. Inverse-trigonometric identities claim that specific angle outputs are equal, so we need one additional check: is the proposed angle in the inverse function’s principal range? This lesson makes that check part of every simplification rather than leaving it implicit.

For example, sin(2θ) = 2 sin θ cos θ is true for every real θ. But sin⁻¹(sin(2θ)) = 2θ is true directly only when −π/2 ≤ 2θ ≤ π/2. If 2θ lies outside that interval, the sine is unchanged by replacing it with an appropriate principal angle, and inverse sine returns that replacement.

The general inverse-composition distinctionLaTeX
F stands for one of the six trigonometric functions with its selected bijective restriction. The restriction for θ is exactly the principal output range of F⁻¹.
Function FRestriction for F(F⁻¹(x))Restriction for direct F⁻¹(F(θ)) cancellation
sin−1 ≤ x ≤ 1−π/2 ≤ θ ≤ π/2
cos−1 ≤ x ≤ 10 ≤ θ ≤ π
cosec|x| ≥ 1−π/2 ≤ θ ≤ π/2 and θ ≠ 0
sec|x| ≥ 10 ≤ θ ≤ π and θ ≠ π/2
tanx ∈ ℝ−π/2 < θ < π/2
cotx ∈ ℝ0 < θ < π

A reliable proof starts by naming the inverse angle, such as θ = sin⁻¹ x. Write both its ordinary trigonometric equation and its range. Then transform the expression, taking care with square roots. Finally check the angle that will be passed through the outer inverse. These steps explain both the result and the input bounds under which it holds.

Double angles and the sign of a square root

An expression containing x√(1 − x²) suggests substituting x = sin θ or x = cos θ. The square root is the nonnegative root, so √(cos² θ) equals |cos θ|, not automatically cos θ. Choosing θ as a principal angle gives the sign information needed to remove that absolute value.

Example — A double inverse-sine identity

Problem
Prove sin⁻¹(2x√(1 − x²)) = 2sin⁻¹x for −1/√2 ≤ x ≤ 1/√2.

  1. 1.Let θ = sin⁻¹ x. Since sine increases on its principal interval, the stated bound gives −π/4 ≤ θ ≤ π/4.
  2. 2.Now x = sin θ. Cosine is nonnegative on this θ interval, so √(1 − x²) = √(1 − sin² θ) = |cos θ| = cos θ.
  3. 3.The expression inside the inverse becomes 2 sin θ cos θ = sin(2θ).
  4. 4.Doubling the θ bounds gives −π/2 ≤ 2θ ≤ π/2. Thus the angle 2θ is already in the principal inverse-sine range.
  5. 5.Therefore sin⁻¹(sin(2θ)) = 2θ = 2 sin⁻¹ x. The argument of inverse sine is also automatically in [−1, 1], since it has been written as a sine.
Check the angle after multiplying it−π/2−π/40π/4π/2θ = sin⁻¹ x−π/2−π/40π/4π/22θDoubling this θ interval stays inside the inverse-sine principal range.The bound |x| ≤ 1/√2 makes direct cancellation valid.
Why the double-angle bound matters— The top interval controls the named angle θ. The bottom interval controls the doubled angle passed through the outer inverse.
Example — A double inverse-cosine expression

Problem
Prove sin⁻¹(2x√(1 − x²)) = 2cos⁻¹x for 1/√2 ≤ x ≤ 1.

  1. 1.Let θ = cos⁻¹ x. Since x is between 1/√2 and 1, the principal cosine angle satisfies 0 ≤ θ ≤ π/4.
  2. 2.Now x = cos θ and sin θ ≥ 0. Therefore √(1 − x²) = √(sin² θ) = sin θ.
  3. 3.The inner expression is 2 cos θ sin θ = sin(2θ).
  4. 4.The angle 2θ lies in [0, π/2], which is inside the inverse-sine principal range.
  5. 5.Thus sin⁻¹(sin(2θ)) = 2θ = 2 cos⁻¹ x. At x = 1, both sides are 0; the included endpoints cause no exception.
Common mistake

The first double-angle identity is not valid merely because |x| ≤ 1. For x = √3/2, its left side is sin⁻¹(√3/2) = π/3, while 2 sin⁻¹ x = 2π/3. The doubled angle has left the principal inverse-sine range.

Triple angles use the same reasoning

The triple-angle formulas create polynomial expressions that look more complicated than the original ratio. Naming the principal angle turns those polynomials back into a sine or cosine of three times that angle. The given bounds ensure that tripling stays in the appropriate principal range.

Ordinary triple-angle identitiesLaTeX
The superscript 3 here is a cube of a ratio; 3θ is three times the angle. The formulas do not themselves select inverse branches.
Example — Triple inverse-sine and inverse-cosine identities

Problem
Prove 3sin⁻¹x = sin⁻¹(3x − 4x³) for x ∈ [−1/2, 1/2], and 3cos⁻¹x = cos⁻¹(4x³ − 3x) for x ∈ [1/2, 1].

  1. 1.For the sine statement, let θ = sin⁻¹ x. The input bound gives −π/6 ≤ θ ≤ π/6, so −π/2 ≤ 3θ ≤ π/2.
  2. 2.The triple-sine formula gives 3x − 4x³ = sin(3θ). Its principal inverse is therefore 3θ, proving the first statement.
  3. 3.For the cosine statement, let φ = cos⁻¹ x. The input bound gives 0 ≤ φ ≤ π/3, so 0 ≤ 3φ ≤ π.
  4. 4.The triple-cosine formula gives 4x³ − 3x = cos(3φ). Its principal inverse is 3φ, proving the second statement.
  5. 5.The two input intervals differ because inverse sine and inverse cosine have different principal output intervals.

Half-angle transformations of trigonometric ratios

A quotient involving 1 ± sin x or 1 ± cos x often simplifies when written with half angles. The target is a familiar tangent or cotangent of a transformed angle. We must still check denominators before cancelling and the transformed angle before removing the inverse.

Example — A ratio containing 1 − sin x

Problem
Simplify tan⁻¹((cos x)/(1 − sin x)) for −3π/2 < x < π/2.

  1. 1.Write c = cos(x/2) and s = sin(x/2). Then cos x = c² − s² = (c − s)(c + s), while 1 − sin x = c² + s² − 2cs = (c − s)².
  2. 2.On the stated open interval, sin x never equals 1. Thus c − s ≠ 0, and the quotient simplifies to (c + s)/(c − s).
  3. 3.Let α = π/4 + x/2. The angle-addition formulas give sin α = (c + s)/√2 and cos α = (c − s)/√2. Hence the quotient is tan α.
  4. 4.The input bounds give −π/2 < α < π/2. Therefore tan⁻¹(tan α) = α.
  5. 5.The simplest form is π/4 + x/2. This derivation does not divide by cos(x/2), which vanishes at the allowed input x = −π.
Example — A cosine half-angle radical

Problem
Simplify tan⁻¹(√((1 − cos x)/(1 + cos x))) for 0 < x < π.

  1. 1.Let θ = x/2. The given interval implies 0 < θ < π/2, so sin θ and cos θ are both positive.
  2. 2.Using 1 − cos x = 2 sin² θ and 1 + cos x = 2 cos² θ, the radical becomes √(tan² θ) = |tan θ|.
  3. 3.Here tan θ > 0, so the radical equals tan θ. The denominator is positive because cos θ ≠ 0.
  4. 4.Since θ is inside the inverse-tangent range, tan⁻¹(tan θ) = θ = x/2.
Example — A shifted tangent ratio

Problem
Simplify tan⁻¹((cos x − sin x)/(cos x + sin x)) for −π/4 < x < 3π/4.

  1. 1.Let α = π/4 − x. The addition formulas give sin α = (cos x − sin x)/√2 and cos α = (cos x + sin x)/√2.
  2. 2.Thus the quotient is tan α whenever its denominator is nonzero.
  3. 3.The stated x interval gives −π/2 < α < π/2. Therefore cos α > 0; the denominator is nonzero and direct inverse cancellation is justified.
  4. 4.The expression equals π/4 − x. This argument remains valid at x = π/2; it does not require dividing numerator and denominator by cos x.

Radicals can be simplified by naming an inverse angle

When an expression contains √(1 + x²), √(a² − x²) or √(x² − 1), a trigonometric substitution can convert the radical into a side ratio. The substitution is useful only if its angle interval fixes the sign of that ratio. We will state those intervals explicitly.

Example — A radical quotient and rationalisation

Problem
Simplify tan⁻¹((√(1 + x²) − 1)/x) for x ≠ 0.

  1. 1.Multiply numerator and denominator by √(1 + x²) + 1. Since (√(1 + x²))² − 1 = x², the quotient becomes x/(√(1 + x²) + 1).
  2. 2.Let θ = tan⁻¹ x. Then −π/2 < θ < π/2, so cos θ > 0 and sec θ = √(1 + tan² θ) = √(1 + x²).
  3. 3.The quotient equals tan θ/(sec θ + 1) = sin θ/(1 + cos θ) = tan(θ/2). This half-angle identity also works for negative θ.
  4. 4.The angle θ/2 lies in (−π/4, π/4), hence in the inverse-tangent principal range.
  5. 5.The simplest form is (1/2) tan⁻¹ x. The original quotient is undefined at x = 0, even though its rationalised extension has a value there.
Example — A scaled inverse-sine substitution

Problem
Simplify tan⁻¹(x/√(a² − x²)) when a > 0 and |x| < a.

  1. 1.Let θ = sin⁻¹(x/a). The strict input bound gives −π/2 < θ < π/2, so cos θ > 0.
  2. 2.Now x = a sin θ and √(a² − x²) = a√(1 − sin² θ) = a cos θ. The assumption a > 0 is needed for this positive-root step.
  3. 3.The quotient is (a sin θ)/(a cos θ) = tan θ.
  4. 4.Direct cancellation is valid on this θ interval. The result is θ = sin⁻¹(x/a). At |x| = a the original denominator is zero, so the strict bound matters.
Example — Connecting inverse cotangent and inverse secant

Problem
Simplify cot⁻¹(1/√(x² − 1)) for x > 1.

  1. 1.Let θ = sec⁻¹ x. Because x > 1, the principal secant angle lies strictly between 0 and π/2.
  2. 2.We have x = sec θ and √(x² − 1) = √(sec² θ − 1) = √(tan² θ).
  3. 3.Tangent is positive on (0, π/2), so the square root equals tan θ. The reciprocal in the expression is therefore cot θ.
  4. 4.The angle θ also lies in the principal cotangent range (0, π). Hence cot⁻¹(cot θ) = θ = sec⁻¹ x.
  5. 5.At x = 1 the original radical denominator vanishes. For x < −1, the principal secant angle has negative tangent, so the same simplification cannot be copied without changing the reasoning.
θ1√(x² − 1)xx = sec θ with 0 < θ < π/2Schematic: use the labelled lengths, not the drawing’s scale.
The secant substitution as a right triangle— For x > 1, use an acute angle with adjacent side 1 and hypotenuse x. The positive opposite side explains why √(x² − 1) equals tan θ.

Triple tangent and nested expressions

The same range-first method applies when a rational expression hides tan(3θ), or when one inverse expression sits inside another function. Work from the innermost operation outwards. If an angle sum is involved, compute its trigonometric ratios using addition formulas and check any denominator that appears.

Ordinary tangent identitiesLaTeX
These forms require nonzero displayed denominators and defined tangents. An inverse of either ratio additionally requires a check on 3θ or α + β.
Example — A rational triple-angle form

Problem
Simplify tan⁻¹((3a²x − x³)/(a³ − 3ax²)) for a > 0 and −a/√3 < x < a/√3.

  1. 1.Let θ = tan⁻¹(x/a). The stated x interval gives −π/6 < θ < π/6.
  2. 2.Divide the ratio’s numerator and denominator by a³. It becomes [3(x/a) − (x/a)³]/[1 − 3(x/a)²], which equals tan(3θ).
  3. 3.Because |x/a| < 1/√3, the denominator 1 − 3(x/a)² is positive and nonzero.
  4. 4.Also −π/2 < 3θ < π/2, so tan⁻¹(tan(3θ)) = 3θ.
  5. 5.The result is 3 tan⁻¹(x/a), with the strict input bounds retained.
Example — Combining a sine double angle and a cosine double angle

Problem
Evaluate tan((1/2)[sin⁻¹(2x/(1 + x²)) + cos⁻¹((1 − y²)/(1 + y²))]) when |x| < 1, y > 0 and xy < 1.

  1. 1.Let α = tan⁻¹ x and β = tan⁻¹ y. Then −π/4 < α < π/4 and 0 < β < π/2.
  2. 2.The double-angle identities give sin(2α) = 2x/(1 + x²) and cos(2β) = (1 − y²)/(1 + y²).
  3. 3.Since 2α is in (−π/2, π/2) and 2β is in (0, π), the two inverse values equal 2α and 2β respectively.
  4. 4.The outer angle is therefore α + β. Its cosine is cos α cos β(1 − xy), which is positive by the stated assumptions, so its tangent is defined.
  5. 5.Apply the tangent addition formula to obtain (x + y)/(1 − xy). The condition xy < 1 makes its denominator positive and excludes a tangent pole.
Example — A tangent of two nonstandard inverse values

Problem
Evaluate tan(sin⁻¹(3/5) + cot⁻¹(3/2)).

  1. 1.Let α = sin⁻¹(3/5). It is acute, so cos α = 4/5 and tan α = 3/4.
  2. 2.Let β = cot⁻¹(3/2). Its positive input gives an acute angle, so tan β = 2/3.
  3. 3.The tangent addition denominator is 1 − (3/4)(2/3) = 1/2, which is nonzero.
  4. 4.The numerator is 3/4 + 2/3 = 17/12. Therefore tan(α + β) = (17/12)/(1/2) = 17/6.
  5. 5.Neither inverse angle needs a familiar special-angle name. Their sine, cosine or tangent ratios provide all the required information.

When an inner angle lies outside the principal range

An expression such as sin⁻¹(sin θ) is still defined for every real θ because sin θ is always an allowed ratio. What can fail is the claim that its output equals θ. Replace θ by an angle with the same ratio in the outer inverse’s principal interval.

Example — Correcting three out-of-range angles

Problem
Evaluate sin⁻¹(sin(2π/3)), tan⁻¹(tan(3π/4)), and cos⁻¹(cos(7π/6)).

  1. 1.For sine, 2π/3 is outside [−π/2, π/2]. Its supplementary angle π/3 has the same sine and lies in that interval. The first result is π/3.
  2. 2.For tangent, subtract one period π from 3π/4. The resulting −π/4 lies in (−π/2, π/2) and has the same tangent. The second result is −π/4.
  3. 3.For cosine, use cos(2π − θ) = cos θ. The angle 2π − 7π/6 = 5π/6 lies in [0, π]. The third result is 5π/6.
  4. 4.Each correction preserves the ordinary ratio while moving its representative angle into the required principal range.

Check your understanding

The correct answer must survive both the algebra and the principal-range check. Some questions ask about a missing assumption rather than a numerical result, because that assumption is what makes a proof valid.

Quiz

Quick check

For θ = sin⁻¹ x, what extra bound makes sin⁻¹(sin(2θ)) = 2θ valid directly?

Quick check

Which statement about a real square root is always true?

Quick check

For 0 < x < π, tan⁻¹ √((1 − cos x)/(1 + cos x)) equals:

Quick check

What is tan⁻¹(tan(3π/4))?

Quick check

For x > 1, cot⁻¹(1/√(x² − 1)) equals:

Quick check

Evaluate tan⁻¹[2 cos(2 sin⁻¹(1/2))].

Quick check

Which restriction justifies 3 cos⁻¹ x = cos⁻¹(4x³ − 3x)?

Quick check

What is sin(π/3 − sin⁻¹(−1/2))?

Quick check

For a > 0 and |x| < a, tan⁻¹(x/√(a² − x²)) equals:

Quick check

What is tan⁻¹(√3) − cot⁻¹(−√3)?

Practice Problems

Practice Problems
  1. Prove the double inverse-sine identity with x ∈ [−1/√2, 1/√2], stating why the square root is cos θ and why 2θ can be cancelled.
  2. Prove the triple inverse-cosine identity for x ∈ [1/2, 1], and explain what happens to the angle bound when it is multiplied by 3.
  3. Simplify tan⁻¹((√(1 + x²) − 1)/x) for x ≠ 0. Explain why a formula valid after rationalisation does not restore the original missing input 0.
  4. Simplify tan⁻¹((cos x − sin x)/(cos x + sin x)) for −π/4 < x < 3π/4 without dividing by cos x.
  5. For a > 0 and −a/√3 < x < a/√3, simplify tan⁻¹((3a²x − x³)/(a³ − 3ax²)). Include the angle and denominator checks.
  6. Evaluate tan⁻¹[2 cos(2 sin⁻¹(1/2))] by working from the innermost inverse outwards.
  7. Evaluate sin⁻¹(sin(2π/3)), tan⁻¹(tan(3π/4)), and cos⁻¹(cos(7π/6)), identifying the principal representative each time.
  8. For |x| < 1, y > 0 and xy < 1, simplify the tangent of half the sum of sin⁻¹(2x/(1 + x²)) and cos⁻¹((1 − y²)/(1 + y²)).
  9. Evaluate tan(sin⁻¹(3/5) + cot⁻¹(3/2)) using side ratios rather than approximate angles.
  10. Explain why replacing √(x² − 1) with tan(sec⁻¹ x) is valid for x > 1 but fails in sign for x < −1.

Key Takeaways

Key Takeaways

• Ordinary trigonometric equality does not by itself justify inverse cancellation; the candidate angle must lie in the principal range. • Name the inverse angle, transform the expression, check signs and denominators, and then check the outer inverse’s range. • A square root of a square is an absolute value; the chosen angle interval determines whether it can be replaced by the signed ratio. • Double- and triple-angle inverse identities have input bounds because angle multiplication can leave the principal interval. • Half-angle formulas and rationalisation reveal simple inverse expressions, but original excluded inputs remain excluded. • Nested expressions are evaluated from inside out; nonstandard angles can be handled exactly using their trigonometric ratios. • For out-of-range inner angles, select a principal angle with the same sine, cosine or tangent instead of cancelling mechanically.

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The Remaining Principal Branches and Values

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Chapter Summary and Practice