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Inverse Trigonometric Functions: Class 12 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A ship's navigation system uses inverse trigonometric functions to determine the bearing angle of a signal. The system computes the following expression to find the final bearing θ (in radians):

θ = sin⁻¹(sin(7π/6)) + cos⁻¹(cos(−5π/4)) + 2tan⁻¹(tan(3π/4))

A ship's navigation system uses inverse trigonometric functions to determine the bearing angle of a signal. The system computes the following expression to find the final bearing θ (in radians):

θ = sin⁻¹(sin(7π/6)) + cos⁻¹(cos(−5π/4)) + 2tan⁻¹(tan(3π/4))

(i) Find the value of sin⁻¹(sin(7π/6)). [1]
(ii) Find the value of cos⁻¹(cos(−5π/4)). [1]
(iii) Find the final bearing angle θ computed by the navigation system. [2]

Show answer
(i) The principal-value branch of sin⁻¹ is [−π/2, π/2].

Since 7π/6 ∉ [−π/2, π/2], we first rewrite:

sin(7π/6) = sin(π + π/6) = −sin(π/6) = −1/2

∴ sin⁻¹(sin(7π/6)) = sin⁻¹(−1/2) = −π/6

(ii) The principal-value branch of cos⁻¹ is [0, π].

Using the property cos⁻¹(−x) = π − cos⁻¹(x) and cos(5π/4) = cos(π + π/4) = −1/√2:

cos⁻¹(cos(−5π/4)) = cos⁻¹(cos(5π/4)) [∵ cos is an even function]

cos(5π/4) = −1/√2

∴ cos⁻¹(cos(−5π/4)) = cos⁻¹(−1/√2) = π − cos⁻¹(1/√2) = π − π/4 = 3π/4

(iii) The principal-value branch of tan⁻¹ is (−π/2, π/2).

Since 3π/4 ∉ (−π/2, π/2), we rewrite:

tan(3π/4) = tan(π − π/4) = −tan(π/4) = −1

∴ tan⁻¹(tan(3π/4)) = tan⁻¹(−1) = −π/4

Substituting the results from (i), (ii) and the above:

θ = sin⁻¹(sin(7π/6)) + cos⁻¹(cos(−5π/4)) + 2·tan⁻¹(tan(3π/4))

θ = (−π/6) + (3π/4) + 2·(−π/4)

θ = −π/6 + 3π/4 − π/2

Taking LCM = 12:

θ = −2π/12 + 9π/12 − 6π/12

θ = (−2 + 9 − 6)π/12

θ = π/12

∴ The final bearing angle computed by the navigation system is θ = π/12.
Q2Case-based4 marks

A navigation app models the bearing angle a ship makes with the east direction using the function f(x) = tan⁻¹(x), where x is the slope of the ship's path. The app needs to evaluate and simplify various inverse trigonometric expressions to display correct bearing angles to the captain.

A navigation app models the bearing angle a ship makes with the east direction using the function

f(x) = tan⁻¹(x), where x is the slope of the ship's path.

Based on this context, answer the following questions:

(i) State the domain and range (principal value branch) of f(x) = tan⁻¹(x). [1 mark]

(ii) The slope of the ship's path at a certain moment is x = 1/√3. Find the bearing angle f(1/√3). [1 mark]

(iii) The navigation system needs to display a simplified bearing angle for the expression

tan⁻¹ ( (1 + x)/(1 − x) ), where x = tan θ and 0 < θ < π/4.

Simplify this expression in terms of θ. [2 marks]

Show answer
(i) For f(x) = tan⁻¹x:

Domain: ℝ (all real numbers, i.e., x ∈ (−∞, ∞))

Range (principal value branch): (−π/2, π/2)

∴ Domain = ℝ and Range = (−π/2, π/2).

(ii) We need to find tan⁻¹(1/√3).

We know that tan(π/6) = 1/√3, and π/6 ∈ (−π/2, π/2).

∴ f(1/√3) = tan⁻¹(1/√3) = π/6.

(iii) Let I = tan⁻¹( (1 + x)/(1 − x) ), where x = tan θ and 0 < θ < π/4.

Substituting x = tan θ,

I = tan⁻¹( (1 + tan θ)/(1 − tan θ) )

Using the standard identity tan(π/4 + θ) = (1 + tan θ)/(1 − tan θ),

I = tan⁻¹( tan(π/4 + θ) )

Since 0 < θ < π/4, we have π/4 < π/4 + θ < π/2, so π/4 + θ ∈ (−π/2, π/2).

∴ I = π/4 + θ.

∴ tan⁻¹( (1 + x)/(1 − x) ) = π/4 + θ.
Q3Case-based4 marks

A satellite navigation system uses angle computations to determine positions. An engineer models a positioning formula involving inverse trigonometric functions. While validating the model, she encounters the expression f(x) = sin⁻¹(2x√(1 − x²)), where x ∈ [−1, 1]. She needs to analyse this function carefully to understand its behaviour in different sub-intervals of its domain, and to simplify and differentiate it for use in her algorithm.

A satellite navigation system uses angle computations to determine positions. An engineer models a positioning formula involving inverse trigonometric functions. While validating the model, she encounters the expression:

f(x) = sin⁻¹(2x√(1 − x²)), where x ∈ [−1, 1]

She needs to analyse this function carefully to understand its behaviour in different sub-intervals of its domain, and to simplify and differentiate it for use in her algorithm.

(i) For x ∈ [1/√2, 1], express f(x) = sin⁻¹(2x√(1 − x²)) in its simplified form (in terms of sin⁻¹x or cos⁻¹x). [1]

(ii) For x ∈ [−1/√2, 1/√2], express f(x) = sin⁻¹(2x√(1 − x²)) in its simplified form. [1]

(iii) Using the simplified form from part (ii), find f′(x) for x ∈ (−1/√2, 1/√2). Hence evaluate f′(1/2). [2]

OR

(iii) The engineer also encounters the expression g(x) = sin⁻¹(√(1 − x²)) + cos⁻¹x for x ∈ (0, 1). Simplify g(x) and hence find g′(x). [2]

Show answer
(i) For x ∈ [1/√2, 1], put x = sin θ, where θ ∈ [π/4, π/2].

∴ 2x√(1 − x²) = 2 sin θ · cos θ = sin 2θ

Now, since θ ∈ [π/4, π/2], we have 2θ ∈ [π/2, π].

For 2θ ∈ [π/2, π]: sin⁻¹(sin 2θ) = π − 2θ (since sin⁻¹ has range [−π/2, π/2])

⟹ f(x) = sin⁻¹(sin 2θ) = π − 2θ = π − 2 sin⁻¹x

∴ f(x) = π − 2 sin⁻¹x for x ∈ [1/√2, 1].

(ii) For x ∈ [−1/√2, 1/√2], put x = sin θ, where θ ∈ [−π/4, π/4].

∴ 2θ ∈ [−π/2, π/2], so sin⁻¹(sin 2θ) = 2θ.

⟹ f(x) = sin⁻¹(sin 2θ) = 2θ = 2 sin⁻¹x

∴ f(x) = 2 sin⁻¹x for x ∈ [−1/√2, 1/√2].

(iii) From part (ii), f(x) = 2 sin⁻¹x for x ∈ (−1/√2, 1/√2).

Differentiating both sides w.r.t. x,

f′(x) = 2 · 1/√(1 − x²) = 2/√(1 − x²)

At x = 1/2:

f′(1/2) = 2/√(1 − 1/4) = 2/√(3/4) = 2/(√3/2) = 4/√3 = 4√3/3

∴ f′(x) = 2/√(1 − x²) and f′(1/2) = 4√3/3.

OR

(iii) For x ∈ (0, 1), put x = cos θ, where θ ∈ (0, π/2).

∴ √(1 − x²) = √(1 − cos²θ) = sin θ (positive since θ ∈ (0, π/2))

g(x) = sin⁻¹(sin θ) + cos⁻¹(cos θ)

Since θ ∈ (0, π/2) ⊂ [−π/2, π/2]: sin⁻¹(sin θ) = θ

Since θ ∈ (0, π/2) ⊂ [0, π]: cos⁻¹(cos θ) = θ

⟹ g(x) = θ + θ = 2θ = 2 cos⁻¹x

∴ g(x) = 2 cos⁻¹x for x ∈ (0, 1).

Differentiating both sides w.r.t. x,

g′(x) = 2 · (−1/√(1 − x²)) = −2/√(1 − x²)

∴ g(x) = 2 cos⁻¹x and g′(x) = −2/√(1 − x²).
Q4Case-based4 marks

A surveyor is mapping two observation points P and Q on a straight coastline. The angles of elevation of a lighthouse L from P and Q are recorded as α = sin⁻¹(3/5) and β = cos⁻¹(5/13), where both angles lie in the principal-value branch of their respective inverse trigonometric functions.

A surveyor is mapping two observation points P and Q on a straight coastline. The angles of elevation of a lighthouse L from P and Q are recorded as α = sin⁻¹(3/5) and β = cos⁻¹(5/13) respectively, where both angles lie in the principal-value branch of their respective inverse trigonometric functions.

(i) Find the value of tan α. [1 mark]
(ii) Find the value of tan β. [1 mark]
(iii) The surveyor needs to compute the combined bearing angle θ = α + β to calibrate the navigation system. Find the exact value of tan(α + β), and hence determine θ. [2 marks]

Show answer
(i) Since α = sin⁻¹(3/5), we have sin α = 3/5, with α ∈ [−π/2, π/2].

Using the identity cos²α = 1 − sin²α,

cos α = √(1 − 9/25) = √(16/25) = 4/5 (positive, since α ∈ [0, π/2])

∴ tan α = sin α / cos α = (3/5)/(4/5) = 3/4

(ii) Since β = cos⁻¹(5/13), we have cos β = 5/13, with β ∈ [0, π].

Using the identity sin²β = 1 − cos²β,

sin β = √(1 − 25/169) = √(144/169) = 12/13 (positive, since β ∈ [0, π])

∴ tan β = sin β / cos β = (12/13)/(5/13) = 12/5

(iii) Using the addition formula for tangent,

tan(α + β) = (tan α + tan β) / (1 − tan α · tan β)

Substituting tan α = 3/4 and tan β = 12/5,

tan α + tan β = 3/4 + 12/5 = 15/20 + 48/20 = 63/20

tan α · tan β = (3/4)(12/5) = 36/20 = 9/5

1 − tan α · tan β = 1 − 9/5 = −4/5

∴ tan(α + β) = (63/20) / (−4/5) = (63/20) × (−5/4) = −63/16

Since tan α = 3/4 > 0 and tan β = 12/5 > 0, both α and β are in (0, π/2), so α + β ∈ (0, π).

∵ tan(α + β) = −63/16 < 0 and α + β ∈ (0, π), we conclude α + β ∈ (π/2, π).

∴ θ = α + β = π − tan⁻¹(63/16)
Q5Case-based4 marks

A navigation engineer encodes directional bearings using inverse trigonometric functions. Three bearing computations must be verified for correctness before the system is deployed. Each computation involves a key property of inverse trigonometric functions and their principal-value branches.

A navigation system uses angles encoded as inverse trigonometric expressions to determine bearings. An engineer needs to verify and simplify the following bearing computations:

(i) The system computes the expression sin⁻¹(sin(7π/6)). Find its correct principal value. (1 mark)

(ii) A calibration check requires evaluating tan⁻¹(1) + cos⁻¹(−1/2). Find the value. (1 mark)

(iii) The system flags a bearing angle θ satisfying

2 tan⁻¹(cos θ) = tan⁻¹(2 cosec θ), θ ≠ 0.

Find all valid values of θ in the principal range (0, π). (2 marks)

Show answer
(i) Principal-value branch of sin⁻¹ is [−π/2, π/2].

Now, 7π/6 ∉ [−π/2, π/2], so we must bring sin(7π/6) into the principal range.

sin(7π/6) = sin(π + π/6) = −sin(π/6) = −1/2.

∴ sin⁻¹(sin(7π/6)) = sin⁻¹(−1/2) = −π/6.

∴ The required principal value is −π/6.

(ii) tan⁻¹(1) = π/4 [∵ tan(π/4) = 1, and π/4 ∈ (−π/2, π/2)].

cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3 [using cos⁻¹(−x) = π − cos⁻¹x].

∴ tan⁻¹(1) + cos⁻¹(−1/2) = π/4 + 2π/3 = 3π/12 + 8π/12 = 11π/12.

∴ The required value is 11π/12.

(iii) Given: 2 tan⁻¹(cos θ) = tan⁻¹(2 cosec θ), θ ∈ (0, π).

Applying the double-angle formula for inverse tangent,

tan⁻¹((2 cos θ)/(1 − cos²θ)) = tan⁻¹(2 cosec θ) ...(i)

[using 2 tan⁻¹x = tan⁻¹(2x/(1 − x²)), valid when |x| < 1, i.e., |cos θ| < 1, i.e., θ ≠ 0, π]

Since 1 − cos²θ = sin²θ, equation (i) becomes

tan⁻¹(2 cos θ / sin²θ) = tan⁻¹(2 / sin θ).

Applying tan to both sides,

2 cos θ / sin²θ = 2 / sin θ.

Multiplying both sides by sin²θ (sin θ ≠ 0 for θ ∈ (0, π)),

2 cos θ = 2 sin θ

⟹ cos θ = sin θ

⟹ tan θ = 1

⟹ θ = π/4 [∵ θ ∈ (0, π) and tan θ = 1 ⟹ θ = π/4].

Verification: |cos(π/4)| = 1/√2 < 1 ✓ (formula condition satisfied).

∴ The required value of θ is π/4.
Q6Case-based4 marks

A surveyor is mapping a hilly terrain. While measuring angles using a digital theodolite, the device displays the angle in a combined inverse-trigonometric expression. The reading on the screen shows:

f(x) = tan⁻¹(sin x / (1 + cos x)), where −π < x < π, x ≠ −π.

The surveyor's assistant claims that this expression simplifies to a single, much simpler form and can then be used to find particular values quickly.

A surveyor is mapping a hilly terrain. While measuring angles using a digital theodolite, the device displays the angle in a combined inverse-trigonometric expression. The reading on the screen shows:

f(x) = tan⁻¹(sin x / (1 + cos x)), where −π < x < π, x ≠ −π.

The surveyor's assistant claims that this expression simplifies to a single, much simpler form and can then be used to find particular values quickly.

(i) Simplify f(x) to its simplest form. [1 mark]
(ii) Using the simplified form, find the value of f(π/3). [1 mark]
(iii) The theodolite also shows a second reading: g = sin[tan⁻¹(tan(5π/4))]. Find the value of g. [2 marks]

Show answer
(i) Simplify f(x) = tan⁻¹(sin x / (1 + cos x)).

Using the identities sin x = 2 sin(x/2) cos(x/2) and 1 + cos x = 2 cos²(x/2),

f(x) = tan⁻¹[ 2 sin(x/2) cos(x/2) / (2 cos²(x/2)) ]

= tan⁻¹[ sin(x/2) / cos(x/2) ]

= tan⁻¹[ tan(x/2) ]

Since −π < x < π, we have −π/2 < x/2 < π/2, which lies in the principal-value branch of tan⁻¹.

∴ f(x) = x/2

(ii) Find f(π/3).

Using the simplified form f(x) = x/2,

f(π/3) = (π/3) / 2

∴ f(π/3) = π/6

(iii) Find g = sin[ tan⁻¹(tan(5π/4)) ].

Step 1 — Bring the argument into the principal-value branch of tan⁻¹.

The principal-value branch of tan⁻¹ is (−π/2, π/2).

5π/4 lies outside this branch. ∵ tan is periodic with period π,

tan(5π/4) = tan(5π/4 − π) = tan(π/4).

∴ tan⁻¹(tan(5π/4)) = tan⁻¹(tan(π/4)) = π/4.

Step 2 — Evaluate the sine.

g = sin[ tan⁻¹(tan(5π/4)) ] = sin(π/4) = 1/√2.

∴ g = 1/√2
Q7Case-based4 marks

A satellite navigation system uses angle encoding. An engineer models a signal angle θ (in radians) using the function f(x) = sin⁻¹(2x√(1 − x²)), x ∈ [−1, 1]. The system requires the function to be split into three sub-domains based on where f(x) simplifies to different standard expressions.

A satellite navigation system uses angle encoding. An engineer models a signal angle θ (in radians) using the function:

f(x) = sin⁻¹(2x√(1 − x²)), x ∈ [−1, 1]

The system requires the function to be split into three sub-domains based on where f(x) simplifies to different standard expressions.

(i) Find the value of f(1/2). [1 mark]
(ii) Find the value of f(−1/√2). [1 mark]
(iii) Show that f(x) = 2sin⁻¹x when x ∈ [−1/√2, 1/√2], and hence find the range of f(x) on this sub-domain. [2 marks]

Show answer
(i) Finding f(1/2):

Put x = 1/2. Since 1/2 ∈ [−1/√2, 1/√2], we use the identity sin⁻¹(2x√(1 − x²)) = 2sin⁻¹x.

∴ f(1/2) = 2sin⁻¹(1/2) = 2 × π/6 = π/3

∴ f(1/2) = π/3

(ii) Finding f(−1/√2):

Put x = −1/√2.

2x√(1 − x²) = 2 × (−1/√2) × √(1 − 1/2) = 2 × (−1/√2) × (1/√2) = 2 × (−1/2) = −1

∴ f(−1/√2) = sin⁻¹(−1) = −π/2

∴ f(−1/√2) = −π/2

(iii) Showing f(x) = 2sin⁻¹x for x ∈ [−1/√2, 1/√2] and finding the range:

Let x = sinθ, where θ ∈ [−π/2, π/2].

For x ∈ [−1/√2, 1/√2], we have sinθ ∈ [−1/√2, 1/√2], so θ ∈ [−π/4, π/4].

Note: 2θ ∈ [−π/2, π/2], which is the principal-value branch of sin⁻¹.

Now, 2x√(1 − x²) = 2sinθ√(1 − sin²θ) = 2sinθ·cosθ = sin2θ

(Here cosθ ≥ 0 since θ ∈ [−π/4, π/4], so √(1 − sin²θ) = cosθ.)

∴ f(x) = sin⁻¹(sin2θ) = 2θ [since 2θ ∈ [−π/2, π/2]]

= 2sin⁻¹x

∴ L.H.S. = R.H.S. Hence proved.

Range of f(x) = 2sin⁻¹x on x ∈ [−1/√2, 1/√2]:

At x = −1/√2: f = 2sin⁻¹(−1/√2) = 2 × (−π/4) = −π/2
At x = 1/√2: f = 2sin⁻¹(1/√2) = 2 × π/4 = π/2

∴ Range of f(x) on [−1/√2, 1/√2] = [−π/2, π/2]
Q8Case-based4 marks

A navigation software uses inverse trigonometric functions to compute turning angles for a drone. The software encodes three angular corrections as:

α = 2 tan⁻¹(1/√3), β = sin⁻¹(sin(5π/6)), γ = cos⁻¹(cos(7π/4))

(i) Find the principal value of α. [1 mark]
(ii) Find the principal value of β. [1 mark]
(iii) The total turning angle programmed by the software is T = α + β + γ. Find the value of T. If a second route requires a turning angle of cos⁻¹(−T/π), find its principal value. [2 marks]

A navigation software uses inverse trigonometric functions to compute turning angles for a drone. The software encodes three angular corrections as:

α = 2 tan⁻¹(1/√3), β = sin⁻¹(sin(5π/6)), γ = cos⁻¹(cos(7π/4))

(i) Find the principal value of α. [1 mark]
(ii) Find the principal value of β. [1 mark]
(iii) The total turning angle programmed by the software is T = α + β + γ. Find the value of T. If a second route requires a turning angle of cos⁻¹(−T/π), find its principal value. [2 marks]

Show answer
(i) The principal value branch of tan⁻¹ is (−π/2, π/2).

tan⁻¹(1/√3) = π/6 ∵ tan(π/6) = 1/√3 and π/6 ∈ (−π/2, π/2)

∴ α = 2 tan⁻¹(1/√3) = 2 × π/6 = π/3

∴ α = π/3

──────────────────────────────────────

(ii) The principal value branch of sin⁻¹ is [−π/2, π/2].

To evaluate sin⁻¹(sin(5π/6)), we first bring 5π/6 into the principal branch.

5π/6 ∉ [−π/2, π/2], so we write:

sin(5π/6) = sin(π − 5π/6) = sin(π/6)

∵ π/6 ∈ [−π/2, π/2],

⟹ sin⁻¹(sin(5π/6)) = sin⁻¹(sin(π/6)) = π/6

∴ β = π/6

──────────────────────────────────────

(iii) The principal value branch of cos⁻¹ is [0, π].

Step 1 — Find γ = cos⁻¹(cos(7π/4)).

7π/4 ∉ [0, π], so we bring it into the principal branch.

cos(7π/4) = cos(2π − π/4) = cos(π/4)

∵ π/4 ∈ [0, π],

⟹ cos⁻¹(cos(7π/4)) = cos⁻¹(cos(π/4)) = π/4

∴ γ = π/4

Step 2 — Compute T.

T = α + β + γ = π/3 + π/6 + π/4

L.C.M. of 3, 6, 4 is 12.

T = 4π/12 + 2π/12 + 3π/12 = 9π/12 = 3π/4

∴ T = 3π/4

Step 3 — Find the principal value of cos⁻¹(−T/π).

−T/π = −(3π/4)/π = −3/4

∵ −3/4 ∈ [−1, 1], cos⁻¹(−3/4) is defined.

Using the identity cos⁻¹(−x) = π − cos⁻¹(x) for x ∈ [−1, 1],

cos⁻¹(−3/4) = π − cos⁻¹(3/4)

∴ The principal value of cos⁻¹(−T/π) = π − cos⁻¹(3/4).
Q9MCQ1 mark

The principal value of sin⁻¹(−1/2) + cos⁻¹(−1/2) is:

Show answer
Option (a) is correct.

Explanation: Using the standard identity sin⁻¹x + cos⁻¹x = π/2 for all x ∈ [−1, 1], with x = −1/2 ∈ [−1, 1],

sin⁻¹(−1/2) + cos⁻¹(−1/2) = π/2.
Q10Short Answer2 marks

Simplify: tan⁻¹(sin x / (1 + cos x)), 0 < x < π.

Show answer
Using the identities sin x = 2 sin(x/2) cos(x/2) and 1 + cos x = 2 cos²(x/2),

tan⁻¹(sin x / (1 + cos x)) = tan⁻¹(2 sin(x/2) cos(x/2) / 2 cos²(x/2))

= tan⁻¹(sin(x/2) / cos(x/2))

= tan⁻¹(tan(x/2))

Since 0 < x < π ⟹ 0 < x/2 < π/2, which lies in the principal range of tan⁻¹, i.e. (−π/2, π/2),

∴ tan⁻¹(sin x / (1 + cos x)) = x/2.
Q11Short Answer2 marks

Simplify: tan⁻¹(cos x / (1 − sin x)), where x ∈ (−π/2, π/2).

Show answer
We write the argument in terms of a known identity.

cos x = 1 − 2sin²(x/2) ... wait, use the half-angle forms directly:

cos x = cos²(x/2) − sin²(x/2) = (cos(x/2) − sin(x/2))(cos(x/2) + sin(x/2))

1 − sin x = 1 − 2sin(x/2)cos(x/2) = (cos(x/2) − sin(x/2))²

∴ cos x / (1 − sin x) = [(cos(x/2) − sin(x/2))(cos(x/2) + sin(x/2))] / (cos(x/2) − sin(x/2))²

= (cos(x/2) + sin(x/2)) / (cos(x/2) − sin(x/2)) ...(i)

Dividing numerator and denominator of (i) by cos(x/2),

= (1 + tan(x/2)) / (1 − tan(x/2)) = tan(π/4 + x/2)

∴ tan⁻¹(cos x / (1 − sin x)) = tan⁻¹(tan(π/4 + x/2))

For x ∈ (−π/2, π/2), we have x/2 ∈ (−π/4, π/4), so π/4 + x/2 ∈ (0, π/2) ⊂ (−π/2, π/2), which is within the principal branch of tan⁻¹.

∴ tan⁻¹(cos x / (1 − sin x)) = π/4 + x/2
Q12Short Answer2 marks

Find the value of sin⁻¹(sin 5π/6).

Show answer
The principal-value branch of sin⁻¹ is [−π/2, π/2].

Since 5π/6 ∉ [−π/2, π/2], we first write sin(5π/6) in a form whose angle lies in the principal range.

sin(5π/6) = sin(π − π/6) = sin(π/6)

∴ sin⁻¹(sin 5π/6) = sin⁻¹(sin π/6) = π/6
Q13Short Answer2 marks

Express tan⁻¹(sin x / (1 + cos x)), where −π < x < π, in the simplest form.

Show answer
We use the half-angle identities: sin x = 2 sin(x/2) cos(x/2) and 1 + cos x = 2 cos²(x/2).

Substituting,

tan⁻¹(sin x / (1 + cos x)) = tan⁻¹(2 sin(x/2) cos(x/2) / 2 cos²(x/2))

= tan⁻¹(sin(x/2) / cos(x/2))

= tan⁻¹(tan(x/2))

Since −π < x < π, we have −π/2 < x/2 < π/2, which lies in the principal-value branch of tan⁻¹, i.e., (−π/2, π/2).

∴ tan⁻¹(sin x / (1 + cos x)) = x/2
Q14Short Answer3 marks

Prove that: tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4).

Show answer
We use the addition formula: tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)), provided xy < 1.

L.H.S. = tan⁻¹(1/2) + tan⁻¹(2/11)

Check the condition: (1/2) × (2/11) = 1/11 < 1, so the formula applies directly.

⟹ L.H.S. = tan⁻¹((1/2 + 2/11) / (1 − (1/2)(2/11)))

Computing the numerator:
1/2 + 2/11 = 11/22 + 4/22 = 15/22

Computing the denominator:
1 − 1/11 = 10/11

⟹ L.H.S. = tan⁻¹((15/22) ÷ (10/11))

⟹ L.H.S. = tan⁻¹((15/22) × (11/10))

⟹ L.H.S. = tan⁻¹(165/220)

⟹ L.H.S. = tan⁻¹(3/4)

= R.H.S.

∴ L.H.S. = R.H.S. Hence proved.
Q15Short Answer3 marks

Prove that: tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4).

Show answer
We use the identity: tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)), provided xy < 1.

L.H.S. = tan⁻¹(1/2) + tan⁻¹(2/11)

Here, x = 1/2 and y = 2/11.

Check: xy = (1/2) × (2/11) = 1/11 < 1, so the addition formula applies directly.

∴ tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹((1/2 + 2/11) / (1 − (1/2)(2/11)))

Numerator: 1/2 + 2/11 = 11/22 + 4/22 = 15/22

Denominator: 1 − 1/11 = 10/11

∴ = tan⁻¹((15/22) ÷ (10/11))

= tan⁻¹((15/22) × (11/10))

= tan⁻¹(165/220)

= tan⁻¹(3/4)

= R.H.S.

∴ L.H.S. = R.H.S. Hence proved.
Q16Short Answer3 marks

Prove that: tan⁻¹(1/4) + tan⁻¹(2/9) = (1/2) cos⁻¹(3/5).

Show answer
We are required to prove that tan⁻¹(1/4) + tan⁻¹(2/9) = (1/2)cos⁻¹(3/5).

Step 1: Simplify the L.H.S. using the addition formula for tan⁻¹.

Since (1/4)(2/9) = 2/36 = 1/18 < 1, we apply the formula tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)) when xy < 1.

L.H.S. = tan⁻¹((1/4 + 2/9) / (1 − (1/4)(2/9)))

= tan⁻¹(((9 + 8)/36) / ((36 − 2)/36))

= tan⁻¹(17/36 ÷ 34/36)

= tan⁻¹(17/34)

= tan⁻¹(1/2) …(i)

Step 2: Simplify the R.H.S. using the half-angle identity.

Let cos⁻¹(3/5) = α, so cos α = 3/5, where α ∈ [0, π].

⟹ cos α = 3/5 ⟹ sin α = 4/5 (since α ∈ [0, π], sin α ≥ 0)

⟹ tan α = (sin α)/(cos α) = (4/5)/(3/5) = 4/3

Using the half-angle formula: tan(α/2) = sin α / (1 + cos α)

∴ tan(α/2) = (4/5) / (1 + 3/5) = (4/5) / (8/5) = 4/8 = 1/2

Since α ∈ [0, π], we have α/2 ∈ [0, π/2], so tan(α/2) > 0 is consistent.

∴ R.H.S. = (1/2)cos⁻¹(3/5) = α/2 = tan⁻¹(1/2) …(ii)

Step 3: Conclude.

From (i) and (ii),

L.H.S. = tan⁻¹(1/2) = R.H.S.

∴ L.H.S. = R.H.S. Hence proved.
Q17Short Answer3 marks

Find the domain of f(x) = sin⁻¹(2x² − 1). Hence, find all values of x for which f(x) = π/6.

Also, write the range of sin⁻¹x other than its principal branch.

Show answer
Domain of f(x) = sin⁻¹(2x² − 1):

The principal-value branch of sin⁻¹ is defined for arguments in [−1, 1].

For f(x) = sin⁻¹(2x² − 1) to be defined, we need:

−1 ≤ 2x² − 1 ≤ 1

From the left inequality:
2x² − 1 ≥ −1 ⟹ 2x² ≥ 0 ⟹ x² ≥ 0, which holds for all x ∈ ℝ.

From the right inequality:
2x² − 1 ≤ 1 ⟹ 2x² ≤ 2 ⟹ x² ≤ 1 ⟹ −1 ≤ x ≤ 1.

∴ Domain of f(x) = [−1, 1].

Finding x for which f(x) = π/6:

sin⁻¹(2x² − 1) = π/6

⟹ 2x² − 1 = sin(π/6) = 1/2

⟹ 2x² = 3/2

⟹ x² = 3/4

⟹ x = ±√3/2

∵ Both √3/2 and −√3/2 lie in [−1, 1] (the domain),

∴ x = √3/2 or x = −√3/2.

Range of sin⁻¹x other than its principal branch:

The principal-value branch of sin⁻¹x has range [−π/2, π/2].

Another branch (other than the principal branch) of sin⁻¹x is [π/2, 3π/2].
Q18Short Answer3 marks

Find the value of: tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−√3/2).

Show answer
The principal-value branches are: tan⁻¹ ∈ (−π/2, π/2), cos⁻¹ ∈ [0, π], sin⁻¹ ∈ [−π/2, π/2].

Evaluating each term:

tan⁻¹(1):
tan⁻¹(1) = π/4, since tan(π/4) = 1 and π/4 ∈ (−π/2, π/2). ...(i)

cos⁻¹(−1/2):
Using cos⁻¹(−x) = π − cos⁻¹(x),
cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3, since 2π/3 ∈ [0, π]. ...(ii)

sin⁻¹(−√3/2):
Using sin⁻¹(−x) = −sin⁻¹(x),
sin⁻¹(−√3/2) = −sin⁻¹(√3/2) = −π/3, since −π/3 ∈ [−π/2, π/2]. ...(iii)

Adding from (i), (ii) and (iii),
tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−√3/2) = π/4 + 2π/3 − π/3

= π/4 + π/3

= 3π/12 + 4π/12

= 7π/12

∴ tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−√3/2) = 7π/12.
Q19Short Answer3 marks

Find the domain of the function f(x) = sin⁻¹(2x² − 3). Hence, find the value of x for which f(x) = −π/6. Also, write the range of sin⁻¹x other than its principal branch.

Show answer
Domain of f(x) = sin⁻¹(2x² − 3):

For sin⁻¹(t) to be defined, we need −1 ≤ t ≤ 1.

∴ −1 ≤ 2x² − 3 ≤ 1

Adding 3 throughout,

2 ≤ 2x² ≤ 4

Dividing by 2,

1 ≤ x² ≤ 2

⟹ 1 ≤ |x| ≤ √2

⟹ x ∈ [−√2, −1] ∪ [1, √2]

∴ Domain of f(x) = [−√2, −1] ∪ [1, √2].

Finding x when f(x) = −π/6:

f(x) = −π/6

⟹ sin⁻¹(2x² − 3) = −π/6

The principal value branch of sin⁻¹ is [−π/2, π/2], and −π/6 ∈ [−π/2, π/2], so:

2x² − 3 = sin(−π/6) = −1/2

⟹ 2x² = 3 − 1/2 = 5/2

⟹ x² = 5/4

⟹ x = ±√5/2

Now checking x = ±√5/2 ≈ ±1.118 against the domain [−√2, −1] ∪ [1, √2]:

∵ 1 ≤ √5/2 ≤ √2 (since 1 ≤ 1.118 ≤ 1.414), both values lie in the domain.

∴ x = √5/2 or x = −√5/2.

Range of sin⁻¹x other than its principal branch:

The principal branch of sin⁻¹x has range [−π/2, π/2].

∴ One other branch of sin⁻¹x is [π/2, 3π/2].
Q20Case-based4 marks

A surveyor records angles α = sin⁻¹(3/5) and β = cos⁻¹(5/13) at a triangular plot. She needs to find tan(α + β) to verify her field measurements.

A surveyor is mapping a triangular plot of land. She records the following measurements:

• The angle of elevation from point A to the top of a vertical pole at point B is sin⁻¹(3/5).
• The angle of elevation from point A to the top of a second vertical pole at point C is cos⁻¹(5/13).

To cross-check her calculations, she needs to verify whether the combined angle (α + β), where α = sin⁻¹(3/5) and β = cos⁻¹(5/13), lies in the correct range, and then evaluate tan(α + β).

(i) State the principal-value branch of sin⁻¹ and cos⁻¹. Hence write the range of (α + β). [1 mark]

(ii) Express both α and β as tan⁻¹ values. [1 mark]

(iii) Evaluate tan(α + β) and state whether (α + β) is acute or obtuse. [2 marks]

Show answer
(i) Principal-value branch of sin⁻¹ is [−π/2, π/2] and of cos⁻¹ is [0, π].

Since 3/5 > 0, we have α = sin⁻¹(3/5) ∈ (0, π/2).
Since 5/13 > 0, we have β = cos⁻¹(5/13) ∈ (0, π/2).

∴ (α + β) ∈ (0, π).

(ii) For α = sin⁻¹(3/5):

In a right triangle with sin α = 3/5, the adjacent side = √(5² − 3²) = √(25 − 9) = √16 = 4.

∴ tan α = 3/4, so α = tan⁻¹(3/4).

For β = cos⁻¹(5/13):

In a right triangle with cos β = 5/13, the opposite side = √(13² − 5²) = √(169 − 25) = √144 = 12.

∴ tan β = 12/5, so β = tan⁻¹(12/5).

(iii) Now, tan α = 3/4 and tan β = 12/5.

Check: tan α · tan β = (3/4) × (12/5) = 36/20 = 9/5 > 1.

∵ tan α > 0, tan β > 0 and tan α · tan β > 1, and both α, β ∈ (0, π/2), we apply the addition formula with the understanding that (α + β) ∈ (π/2, π).

Using the addition formula:

tan(α + β) = (tan α + tan β) / (1 − tan α · tan β)

⟹ tan(α + β) = (3/4 + 12/5) / (1 − 9/5)

⟹ tan(α + β) = (15/20 + 48/20) / ((5 − 9)/5)

⟹ tan(α + β) = (63/20) / (−4/5)

⟹ tan(α + β) = (63/20) × (−5/4)

⟹ tan(α + β) = −315/80 = −63/16

∵ tan(α + β) < 0 and (α + β) ∈ (π/2, π),

∴ (α + β) is obtuse.

∴ tan(α + β) = −63/16, and (α + β) is an obtuse angle.
Q21Case-based4 marks

A navigation app uses inverse trigonometric functions to calculate angles of elevation and direction. The app programmer defines a function f(x) = sin⁻¹(x) + cos⁻¹(x) for all x in its domain.

A navigation app uses inverse trigonometric functions to calculate the angle of elevation and direction. The app programmer defines a function f(x) = sin⁻¹(x) + cos⁻¹(x) for all x in its domain.

Based on the above information, answer the following sub-questions:

(i) What is the domain of f(x) = sin⁻¹(x) + cos⁻¹(x)? [1]

(ii) What is the value of f(x) for every x in its domain? Justify your answer. [1]

(iii) The app also computes the expression: sin⁻¹(1) + cos⁻¹(−1/2) + tan⁻¹(1). Find the value of this expression. [2]

OR

(iii) The app evaluates tan(sin⁻¹(√3/2) − cos⁻¹(0)). Find its value. [2]

Show answer
(i) The domain of sin⁻¹(x) is [−1, 1] and the domain of cos⁻¹(x) is [−1, 1].
∴ Domain of f(x) = sin⁻¹(x) + cos⁻¹(x) is [−1, 1].

(ii) Using the standard identity: sin⁻¹(x) + cos⁻¹(x) = π/2, for all x ∈ [−1, 1],
∴ f(x) = π/2 for every x in its domain.

(iii) We evaluate: sin⁻¹(1) + cos⁻¹(−1/2) + tan⁻¹(1).

Using principal-value branches — sin⁻¹ ∈ [−π/2, π/2], cos⁻¹ ∈ [0, π], tan⁻¹ ∈ (−π/2, π/2):

sin⁻¹(1) = π/2

For cos⁻¹(−1/2): using cos⁻¹(−x) = π − cos⁻¹(x),
cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3

tan⁻¹(1) = π/4

∴ sin⁻¹(1) + cos⁻¹(−1/2) + tan⁻¹(1) = π/2 + 2π/3 + π/4
= 6π/12 + 8π/12 + 3π/12
= 17π/12

OR

(iii) We evaluate: tan(sin⁻¹(√3/2) − cos⁻¹(0)).

Using principal-value branches — sin⁻¹ ∈ [−π/2, π/2], cos⁻¹ ∈ [0, π]:

sin⁻¹(√3/2) = π/3

cos⁻¹(0) = π/2

∴ sin⁻¹(√3/2) − cos⁻¹(0) = π/3 − π/2 = 2π/6 − 3π/6 = −π/6

∴ tan(sin⁻¹(√3/2) − cos⁻¹(0)) = tan(−π/6) = −tan(π/6) = −1/√3

∴ Required value = −1/√3 (or equivalently −√3/3).
Q22Case-based4 marks

A surveyor is mapping a triangular plot of land using trigonometric instruments. Several of her calculations involve inverse trigonometric functions. Help her evaluate the following expressions using standard principal-value branch rules.

A surveyor is mapping a triangular plot of land. Using her instruments, she records the following angle data:

(i) She computes the expression sin⁻¹(1) + cos⁻¹(0). What is its value?

(ii) She uses the formula for the sum of two inverse tangents and evaluates tan⁻¹(1) + tan⁻¹(√3). Find its value.

(iii) While recording bearings, she needs to find the value of cos⁻¹(cos(7π/6)). What is the correct value? Also, justify why it is NOT 7π/6.

OR

(iii) The surveyor needs to determine the value of k such that:
sin⁻¹(k · tan(2 cos⁻¹(1/2))) = π/6
Find k.

Show answer
Sub-part (i) [1 mark]

The principal-value branch of sin⁻¹ is [−π/2, π/2] and of cos⁻¹ is [0, π].

sin⁻¹(1) = π/2 and cos⁻¹(0) = π/2

∴ sin⁻¹(1) + cos⁻¹(0) = π/2 + π/2 = π

Sub-part (ii) [1 mark]

The principal-value branch of tan⁻¹ is (−π/2, π/2).

tan⁻¹(1) = π/4 and tan⁻¹(√3) = π/3

Here the product of the arguments = 1 × √3 = √3 > 1, and both arguments are positive.

∴ Using tan⁻¹x + tan⁻¹y = π + tan⁻¹((x + y)/(1 − xy)) when x > 0, y > 0, xy > 1,

tan⁻¹(1) + tan⁻¹(√3) = π + tan⁻¹((1 + √3)/(1 − √3))

= π + tan⁻¹(−(√3 + 1)/(√3 − 1))

Now (√3 + 1)/(√3 − 1) = (√3 + 1)²/((√3)² − 1²) = (4 + 2√3)/2 = 2 + √3 = tan(75°) = tan(5π/12)

∴ tan⁻¹(−(2 + √3)) = −5π/12

∴ tan⁻¹(1) + tan⁻¹(√3) = π − 5π/12 = 7π/12

Alternatively, since tan⁻¹(1) + tan⁻¹(√3) = π/4 + π/3 = 3π/12 + 4π/12 = 7π/12 ✓

∴ tan⁻¹(1) + tan⁻¹(√3) = 7π/12

Sub-part (iii) [2 marks]

The principal-value branch of cos⁻¹ is [0, π].

Now 7π/6 ∉ [0, π], so cos⁻¹(cos(7π/6)) ≠ 7π/6.

First compute cos(7π/6):

cos(7π/6) = cos(π + π/6) = −cos(π/6) = −√3/2

∴ cos⁻¹(cos(7π/6)) = cos⁻¹(−√3/2)

Since cos(5π/6) = −√3/2 and 5π/6 ∈ [0, π],

∴ cos⁻¹(cos(7π/6)) = 5π/6

Justification: cos⁻¹ returns values only in [0, π]. Since 7π/6 > π, it lies outside the principal-value branch. The angle 5π/6 ∈ [0, π] has the same cosine value (−√3/2), so the correct answer is 5π/6.

OR

Sub-part (iii) OR [2 marks]

Given: sin⁻¹(k · tan(2 cos⁻¹(1/2))) = π/6

Step 1: Evaluate cos⁻¹(1/2).

cos⁻¹(1/2) = π/3 ∵ cos(π/3) = 1/2 and π/3 ∈ [0, π]

⟹ 2 cos⁻¹(1/2) = 2π/3

Step 2: Evaluate tan(2π/3).

tan(2π/3) = tan(π − π/3) = −tan(π/3) = −√3

Step 3: Substitute into the given equation.

sin⁻¹(k · (−√3)) = π/6

⟹ k · (−√3) = sin(π/6) = 1/2

⟹ k = −1/(2√3) = −√3/6

∴ k = −√3/6
Q23Case-based4 marks

A surveyor is mapping the elevation angles at two observation posts P and Q on a straight road. At post P, the angle of elevation to a tower top is recorded as sin⁻¹(√3/2), and at post Q, the angle is recorded as cos⁻¹(1/2). A third reading at post R gives the angle as tan⁻¹(1/√3). The surveyor wishes to analyse these inverse trigonometric readings using principal value branches.

A surveyor is mapping the elevation angles at two observation posts P and Q on a straight road. At post P, the angle of elevation to a tower top is recorded as sin⁻¹(√3/2), and at post Q, the angle is recorded as cos⁻¹(1/2). A third reading at post R gives the angle as tan⁻¹(1/√3).

Based on this context, answer the following:

(i) What is the principal value of sin⁻¹(√3/2)? State the principal value branch used. [1]

(ii) Show that the angles recorded at posts P and Q are equal. [1]

(iii) The surveyor notes that the combined reading at posts P and R together equals the reading at post Q. Verify whether the following identity holds for these readings:

sin⁻¹(√3/2) + tan⁻¹(1/√3) = cos⁻¹(1/2) + π/3

OR

(iii) Using the principal values found above, evaluate:

cos⁻¹(1/2) − sin⁻¹(−√3/2) + tan⁻¹(−1/√3) [2]

Show answer
SECTION E — Case Study Answer

(i) [1 mark]

The principal value branch of sin⁻¹ is [−π/2, π/2].

We require θ ∈ [−π/2, π/2] such that sin θ = √3/2.

∴ sin⁻¹(√3/2) = π/3

(ii) [1 mark]

The principal value branch of cos⁻¹ is [0, π].

We require φ ∈ [0, π] such that cos φ = 1/2.

∴ cos⁻¹(1/2) = π/3

From (i) and (ii), sin⁻¹(√3/2) = π/3 = cos⁻¹(1/2).

∴ The angles recorded at posts P and Q are equal. Hence proved.

(iii) Main option [2 marks]

The principal value branch of tan⁻¹ is (−π/2, π/2).

We require α ∈ (−π/2, π/2) such that tan α = 1/√3.

∴ tan⁻¹(1/√3) = π/6

L.H.S. = sin⁻¹(√3/2) + tan⁻¹(1/√3)

= π/3 + π/6

= 2π/6 + π/6

= 3π/6

= π/2

R.H.S. = cos⁻¹(1/2) + π/3

= π/3 + π/3

= 2π/3

∵ π/2 ≠ 2π/3,

∴ L.H.S. ≠ R.H.S.

Hence, the given identity does NOT hold for these readings.

∴ The surveyor's combined reading identity is incorrect; sin⁻¹(√3/2) + tan⁻¹(1/√3) = π/2, which is NOT equal to cos⁻¹(1/2) + π/3 = 2π/3.

─────────────────────────────

(iii) OR option [2 marks]

Using properties of inverse trigonometric functions:

sin⁻¹(−x) = −sin⁻¹(x) and tan⁻¹(−x) = −tan⁻¹(x)

∴ sin⁻¹(−√3/2) = −sin⁻¹(√3/2) = −π/3

∴ tan⁻¹(−1/√3) = −tan⁻¹(1/√3) = −π/6

Now,

cos⁻¹(1/2) − sin⁻¹(−√3/2) + tan⁻¹(−1/√3)

= π/3 − (−π/3) + (−π/6)

= π/3 + π/3 − π/6

= 2π/6 + 2π/6 − π/6

= 3π/6

= π/2

∴ cos⁻¹(1/2) − sin⁻¹(−√3/2) + tan⁻¹(−1/√3) = π/2
Q24Case-based4 marks

A surveyor standing at point O on level ground observes the tops of two vertical towers P and Q standing on the same horizontal line as O. The angle of elevation to the top of tower P is α = tan⁻¹(3/4) and the angle of elevation to the top of tower Q is β = tan⁻¹(5/12). The surveyor wishes to find the combined angular measure α + β and use it to determine certain trigonometric values relevant to his survey calculations.

A surveyor standing at a point O on level ground observes the tops of two vertical towers, P and Q, standing on the same horizontal line as O. The angle of elevation of the top of tower P is tan⁻¹(3/4) and the angle of elevation of the top of tower Q is tan⁻¹(5/12). The surveyor also notes that:

(i) Express tan⁻¹(3/4) + tan⁻¹(5/12) as a single inverse trigonometric value. [1 mark]

(ii) Hence, find the value of sin[tan⁻¹(3/4) + tan⁻¹(5/12)]. [1 mark]

(iii) The surveyor records that the combined angular measure equals cos⁻¹(k). Find the exact value of k. Also verify that sin²[tan⁻¹(3/4) + tan⁻¹(5/12)] + cos²[tan⁻¹(3/4) + tan⁻¹(5/12)] = 1. [2 marks]

Diagram for question 24: Inverse Trigonometric Functions
Show answer
Sub-part (i): [1 mark]

Let α = tan⁻¹(3/4) and β = tan⁻¹(5/12).

Using the addition formula for inverse tangent:
tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)), provided xy < 1.

Here, xy = (3/4)(5/12) = 15/48 = 5/16 < 1.

∴ α + β = tan⁻¹((3/4 + 5/12)/(1 − 5/16))

Numerator: 3/4 + 5/12 = 9/12 + 5/12 = 14/12 = 7/6

Denominator: 1 − 5/16 = 11/16

∴ α + β = tan⁻¹((7/6) ÷ (11/16)) = tan⁻¹((7/6) × (16/11)) = tan⁻¹(112/66) = tan⁻¹(56/33)

∴ tan⁻¹(3/4) + tan⁻¹(5/12) = tan⁻¹(56/33)

─────────────────────────────────────────
Sub-part (ii): [1 mark]

Let θ = tan⁻¹(56/33), so tan θ = 56/33.

Constructing the right triangle: opposite = 56, adjacent = 33.

Hypotenuse = √(56² + 33²) = √(3136 + 1089) = √4225 = 65

Since θ = tan⁻¹(56/33) ∈ (0, π/2), sin θ > 0.

∴ sin[tan⁻¹(3/4) + tan⁻¹(5/12)] = sin θ = 56/65

─────────────────────────────────────────
Sub-part (iii): [2 marks]

From sub-part (ii), with tan θ = 56/33 and hypotenuse = 65:

cos θ = 33/65

Since the combined angle θ = tan⁻¹(56/33) = cos⁻¹(k),

∴ k = cos θ = 33/65

Verification:

sin²θ + cos²θ = (56/65)² + (33/65)²

= 3136/4225 + 1089/4225

= 4225/4225

= 1

∴ sin²[tan⁻¹(3/4) + tan⁻¹(5/12)] + cos²[tan⁻¹(3/4) + tan⁻¹(5/12)] = 1. Hence verified.

∴ The exact value of k = 33/65.
Q25Case-based4 marks

A navigation system encodes directional angles using inverse trigonometric functions. All output values must lie within the principal-value branch of each function for the system to process them correctly. The engineer must bring each angle into the appropriate principal-value range before applying the inverse function.

A navigation system uses angles to determine directions. The system stores angles as values of inverse trigonometric functions. An engineer is analysing the system and encounters the following four expressions:

(i) sin⁻¹(sin 5π/6)
(ii) cos⁻¹(cos 4π/3)
(iii) tan⁻¹(tan 3π/4)
(iv) sin⁻¹(sin 13π/5)

The engineer must simplify each expression to its principal value so that the navigation system can process it correctly.

(a) What is the principal-value range of sin⁻¹x? Evaluate sin⁻¹(sin 5π/6). [1 mark]
(b) Evaluate cos⁻¹(cos 4π/3). [1 mark]
(c) Evaluate tan⁻¹(tan 3π/4) and sin⁻¹(sin 13π/5). [2 marks]

Show answer
(a) The principal-value branch of sin⁻¹x is [−π/2, π/2].

Now, 5π/6 ∉ [−π/2, π/2], so we write:
sin⁻¹(sin 5π/6) = sin⁻¹(sin(π − π/6))
= sin⁻¹(sin π/6) [∵ sin(π − θ) = sin θ]

∵ π/6 ∈ [−π/2, π/2],

∴ sin⁻¹(sin 5π/6) = π/6

(b) The principal-value branch of cos⁻¹x is [0, π].

Now, 4π/3 ∉ [0, π], so we write:
cos⁻¹(cos 4π/3) = cos⁻¹(cos(2π − 2π/3))

∵ cos(2π − θ) = cos θ, this gives cos(2π/3), but let us use a simpler reduction:
cos 4π/3 = cos(π + π/3) = −cos π/3 = −1/2

⟹ cos⁻¹(cos 4π/3) = cos⁻¹(−1/2)

∵ cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3, and 2π/3 ∈ [0, π],

∴ cos⁻¹(cos 4π/3) = 2π/3

(c) The principal-value branch of tan⁻¹x is (−π/2, π/2).

3π/4 ∉ (−π/2, π/2), so:
tan⁻¹(tan 3π/4) = tan⁻¹(tan(π − π/4))
= tan⁻¹(−tan π/4) [∵ tan(π − θ) = −tan θ]
= tan⁻¹(−1)

∵ tan⁻¹(−1) = −π/4 ∈ (−π/2, π/2),

∴ tan⁻¹(tan 3π/4) = −π/4

For sin⁻¹(sin 13π/5):
13π/5 = 2π + 3π/5
⟹ sin(13π/5) = sin(3π/5) = sin(π − 2π/5) = sin(2π/5)

∵ 2π/5 ∈ [−π/2, π/2],

∴ sin⁻¹(sin 13π/5) = 2π/5
Q26Case-based4 marks

A surveyor is determining the angle of elevation to the top of a vertical tower from two different points on the ground. After careful measurements, the surveyor models the combined angle calculation as the expression f(x) = tan⁻¹((1 + sin x)/(cos x)) + tan⁻¹((1 - sin x)/(cos x)), where 0 < x < π/2.

A surveyor is determining the angle of elevation to the top of a vertical tower from two different points on the ground. After careful measurements, the surveyor models the combined angle calculation as the expression

f(x) = tan⁻¹((1 + sin x)/(cos x)) + tan⁻¹((1 - sin x)/(cos x)), where 0 < x < π/2.

(i) Simplify tan⁻¹((1 + sin x)/(cos x)) into the simplest form involving x. [1 mark]
(ii) Simplify tan⁻¹((1 - sin x)/(cos x)) into the simplest form involving x. [1 mark]
(iii) Using parts (i) and (ii), find the value of f(x). Hence determine the value of f(π/3). [2 marks]

Show answer
(i) Simplify tan⁻¹((1 + sin x)/(cos x)):

Using the identities 1 + sin x = (cos(x/2) + sin(x/2))² and cos x = cos²(x/2) − sin²(x/2) = (cos(x/2) + sin(x/2))(cos(x/2) − sin(x/2)),

tan⁻¹((1 + sin x)/(cos x)) = tan⁻¹((cos(x/2) + sin(x/2))² / [(cos(x/2) + sin(x/2))(cos(x/2) − sin(x/2))])

⟹ = tan⁻¹((cos(x/2) + sin(x/2)) / (cos(x/2) − sin(x/2)))

Dividing numerator and denominator by cos(x/2),

= tan⁻¹((1 + tan(x/2)) / (1 − tan(x/2)))

Using tan⁻¹((1 + tan θ)/(1 − tan θ)) = tan⁻¹(tan(π/4 + θ)) = π/4 + θ (valid since 0 < x < π/2 ⟹ 0 < x/2 < π/4, so π/4 + x/2 ∈ (π/4, π/2) ⊂ (−π/2, π/2)),

∴ tan⁻¹((1 + sin x)/(cos x)) = π/4 + x/2

(ii) Simplify tan⁻¹((1 − sin x)/(cos x)):

Using 1 − sin x = (cos(x/2) − sin(x/2))²,

tan⁻¹((1 − sin x)/(cos x)) = tan⁻¹((cos(x/2) − sin(x/2))² / [(cos(x/2) + sin(x/2))(cos(x/2) − sin(x/2))])

⟹ = tan⁻¹((cos(x/2) − sin(x/2)) / (cos(x/2) + sin(x/2)))

Dividing numerator and denominator by cos(x/2),

= tan⁻¹((1 − tan(x/2)) / (1 + tan(x/2)))

Using tan⁻¹((1 − tan θ)/(1 + tan θ)) = tan⁻¹(tan(π/4 − θ)) = π/4 − θ (valid since π/4 − x/2 ∈ (0, π/4) ⊂ (−π/2, π/2)),

∴ tan⁻¹((1 − sin x)/(cos x)) = π/4 − x/2

(iii) Value of f(x) and f(π/3):

From (i) and (ii),

f(x) = tan⁻¹((1 + sin x)/(cos x)) + tan⁻¹((1 − sin x)/(cos x))

= (π/4 + x/2) + (π/4 − x/2)

= π/4 + π/4

∴ f(x) = π/2 for all x ∈ (0, π/2).

Substituting x = π/3,

∴ f(π/3) = π/2.
Q27Case-based4 marks

A software engineer is designing an automated navigation system that uses inverse trigonometric functions to compute steering correction angles. The steering correction angle θ is modelled by the expression θ = tan⁻¹(1/√3) + cos⁻¹(−√3/2) − sin⁻¹(sin(5π/3)). The system flags an error if |θ| > 2.

A software engineer is designing an automated navigation system. The system uses inverse trigonometric functions to determine steering angles. During testing, the engineer models the steering correction angle θ (in radians) using the expression:

θ = tan⁻¹(1/√3) + cos⁻¹(−√3/2) − sin⁻¹(sin(5π/3))

The system flags an error if |θ| > 2. The engineer also needs to verify that the domain condition for the expression cos⁻¹(2x − 1) to be defined restricts x to a specific interval, and that tan⁻¹(x) + tan⁻¹(1/x) simplifies correctly for x < 0, to avoid division-by-zero errors in the code.

(i) State the principal value branch (range) of cos⁻¹(x) and evaluate cos⁻¹(−√3/2). [1 mark]

(ii) Evaluate sin⁻¹(sin(5π/3)), clearly bringing the argument into the principal value branch. [1 mark]

(iii) Compute the full steering correction angle θ = tan⁻¹(1/√3) + cos⁻¹(−√3/2) − sin⁻¹(sin(5π/3)), and determine whether the system flags an error. [2 marks]

Show answer
(i) The principal value branch of cos⁻¹(x) has range [0, π].

Let cos⁻¹(−√3/2) = α, where α ∈ [0, π].

∵ cos(5π/6) = −√3/2 and 5π/6 ∈ [0, π],

∴ cos⁻¹(−√3/2) = 5π/6

(ii) The principal value branch of sin⁻¹(x) has range [−π/2, π/2].

Now, 5π/3 ∉ [−π/2, π/2], so we bring it into the principal range.

sin(5π/3) = sin(2π − π/3) = −sin(π/3) = −√3/2

∵ sin(−π/3) = −√3/2 and −π/3 ∈ [−π/2, π/2],

∴ sin⁻¹(sin(5π/3)) = sin⁻¹(−√3/2) = −π/3

(iii) The principal value branch of tan⁻¹(x) has range (−π/2, π/2).

tan⁻¹(1/√3) = π/6, since tan(π/6) = 1/√3 and π/6 ∈ (−π/2, π/2).

Now substituting all three evaluated values:

θ = tan⁻¹(1/√3) + cos⁻¹(−√3/2) − sin⁻¹(sin(5π/3))

θ = π/6 + 5π/6 − (−π/3)

θ = π/6 + 5π/6 + π/3

θ = (π + 5π)/6 + π/3

θ = 6π/6 + π/3

θ = π + π/3

θ = 3π/3 + π/3

θ = 4π/3

Now, |θ| = 4π/3 ≈ 4.19

∵ 4π/3 > 2,

∴ θ = 4π/3 and |θ| > 2, so the system DOES flag an error.
Q28Case-based4 marks

A software engineer is designing a rotation-animation module that maps angles to principal-value outputs using inverse trigonometric functions. The four test cases listed must each be reduced to a value lying strictly within the principal-value branch of the respective inverse function.

A software engineer is designing a rotation-animation module. The module computes angles using inverse trigonometric functions. While testing, the engineer evaluates the following four expressions:

(i) sin⁻¹(sin(7π/6))
(ii) cos⁻¹(cos(5π/3))
(iii) tan⁻¹(tan(−3π/4))
(iv) sin⁻¹(cos(−π/3))

The engineer must return each result in the principal-value branch of the respective inverse function. Find the value of each expression.

Show answer
Principal-value branches used throughout:
• sin⁻¹ ∈ [−π/2, π/2]
• cos⁻¹ ∈ [0, π]
• tan⁻¹ ∈ (−π/2, π/2)

(i) sin⁻¹(sin(7π/6))

7π/6 ∉ [−π/2, π/2], so we reduce sin(7π/6) first.

sin(7π/6) = sin(π + π/6) = −sin(π/6) = −1/2

∴ sin⁻¹(sin(7π/6)) = sin⁻¹(−1/2)

∵ sin(−π/6) = −1/2 and −π/6 ∈ [−π/2, π/2],

∴ sin⁻¹(sin(7π/6)) = −π/6

(ii) cos⁻¹(cos(5π/3))

5π/3 ∉ [0, π], so we reduce cos(5π/3) first.

cos(5π/3) = cos(2π − π/3) = cos(π/3) = 1/2

∴ cos⁻¹(cos(5π/3)) = cos⁻¹(1/2)

∵ cos(π/3) = 1/2 and π/3 ∈ [0, π],

∴ cos⁻¹(cos(5π/3)) = π/3

(iii) tan⁻¹(tan(−3π/4))

−3π/4 ∉ (−π/2, π/2), so we use the periodicity of tan.

tan(−3π/4) = tan(−3π/4 + π) = tan(π/4)

∴ tan⁻¹(tan(−3π/4)) = tan⁻¹(tan(π/4))

∵ π/4 ∈ (−π/2, π/2),

∴ tan⁻¹(tan(−3π/4)) = π/4

(iv) sin⁻¹(cos(−π/3))

First evaluate cos(−π/3):
cos(−π/3) = cos(π/3) = 1/2

∴ sin⁻¹(cos(−π/3)) = sin⁻¹(1/2)

∵ sin(π/6) = 1/2 and π/6 ∈ [−π/2, π/2],

∴ sin⁻¹(cos(−π/3)) = π/6
Q29Case-based4 marks

A satellite navigation system encodes angular positions using inverse trigonometric functions. An engineer analyses encoded signals of the form sin⁻¹(a), cos⁻¹(b) and tan⁻¹(c) to decode precise angular data. Correct identification of principal value branches and accurate computation of inverse trigonometric expressions are critical to avoid navigation errors.

A satellite navigation system encodes angular positions using inverse trigonometric functions. An engineer is analysing three encoded angles:

(i) Find the domain and range (principal value branch) of f(x) = cos⁻¹x. [1]

(ii) The system receives a signal encoded as sin⁻¹(−1/2) + cos⁻¹(−1/2). Decode this by finding its exact value. [1]

(iii) A transmission error occurs when two encoded values are added incorrectly. The erroneous computation gives:

tan⁻¹(1/2) + tan⁻¹(2/3) = tan⁻¹(x)

Find the correct value of x, and hence determine the exact decoded angle in radians. Verify that the result lies in the principal value branch of tan⁻¹. [2]

Show answer
(i) For f(x) = cos⁻¹x:

Domain: [−1, 1]

Range (principal value branch): [0, π]

∴ Domain = [−1, 1] and Range = [0, π].

(ii) We use the standard results:

sin⁻¹(−1/2) = −π/6 [∵ sin(π/6) = 1/2 and sin⁻¹ ∈ [−π/2, π/2]]

cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3 [∵ cos⁻¹(−x) = π − cos⁻¹x and cos⁻¹ ∈ [0, π]]

∴ sin⁻¹(−1/2) + cos⁻¹(−1/2) = −π/6 + 2π/3 = −π/6 + 4π/6 = 3π/6 = π/2

∴ The decoded angle = π/2.

(iii) We apply the addition formula:

tan⁻¹a + tan⁻¹b = tan⁻¹((a + b)/(1 − ab)), provided ab < 1.

Here a = 1/2, b = 2/3.

ab = (1/2)(2/3) = 1/3 < 1, so the formula applies directly.

tan⁻¹(1/2) + tan⁻¹(2/3) = tan⁻¹((1/2 + 2/3)/(1 − 1/3))

= tan⁻¹((3/6 + 4/6)/(2/3))

= tan⁻¹((7/6) × (3/2))

= tan⁻¹(7/4)

∴ x = 7/4, and the correct decoded angle = tan⁻¹(7/4).

Verification: Since 7/4 > 0, we have tan⁻¹(7/4) ∈ (0, π/2) ⊂ (−π/2, π/2), which is the principal value branch of tan⁻¹.

∴ The decoded angle = tan⁻¹(7/4), which lies in the principal value branch (−π/2, π/2). Hence proved.
Q30Case-based4 marks

A navigation software models the angle of deviation of a ship from its base direction using inverse trigonometric functions. The deviation angle θ is expressed as:

θ = sin⁻¹(√3/2) + cos⁻¹(−1/2) + tan⁻¹(−1/√3)

A navigation software models the angle of deviation of a ship from its base direction using inverse trigonometric functions. The deviation angle θ is expressed as:

θ = sin⁻¹(√3/2) + cos⁻¹(−1/2) + tan⁻¹(−1/√3)

Based on the above information, answer the following questions:

(i) What is the principal value of sin⁻¹(√3/2)? State the principal-value branch used. [1]

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

(iii) Find the total deviation angle θ. Hence determine whether the deviation is less than, equal to, or greater than π radians. [2]

Show answer
(i) The principal-value branch of sin⁻¹ is [−π/2, π/2].

Since sin(π/3) = √3/2 and π/3 ∈ [−π/2, π/2],

∴ sin⁻¹(√3/2) = π/3

(ii) The principal-value branch of cos⁻¹ is [0, π].

Using the identity cos⁻¹(−x) = π − cos⁻¹(x),

cos⁻¹(−1/2) = π − cos⁻¹(1/2)

Since cos(π/3) = 1/2,

cos⁻¹(−1/2) = π − π/3 = 2π/3

∴ cos⁻¹(−1/2) = 2π/3

(iii) The principal-value branch of tan⁻¹ is (−π/2, π/2).

Using the identity tan⁻¹(−x) = −tan⁻¹(x),

tan⁻¹(−1/√3) = −tan⁻¹(1/√3)

Since tan(π/6) = 1/√3,

tan⁻¹(−1/√3) = −π/6

Now substituting all three values,

θ = sin⁻¹(√3/2) + cos⁻¹(−1/2) + tan⁻¹(−1/√3)

θ = π/3 + 2π/3 + (−π/6)

θ = π − π/6

θ = 5π/6

∴ θ = 5π/6

Since 5π/6 < π,

∴ The total deviation angle θ = 5π/6 radians, which is less than π radians.

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