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NDA Maths · Inverse Trigonometry

Evaluating Composite Inverse Expressions

Composite expressions nest a trig function around an inverse one (or vice versa); evaluate from the inside out, naming the inner inverse as an angle and building the right triangle for it.

All Inverse Trigonometry notesNDA Maths strategy

Why this matters

11 PYQs, 4 HARD — the chapter's toughest pocket. The reliable method is always the same: set the innermost inverse equal to an angle θ, read off its sin/cos/tan from a right triangle, then evaluate the outer functions. Double- and half-angle formulas finish the job.

Concept 1 of 3

Inner-to-Outer Evaluation & sin⁻¹(sin x)

Intuition

Peel the expression from the inside: evaluate the innermost inverse to a concrete angle, then apply the outer functions one at a time. For sin⁻¹(sin x) the answer is x ONLY if x is already in the principal range — otherwise reduce it.

Definition

  • Nested evaluation: name the inner inverse θ=csc⁡−12\theta = \csc^{-1}2 (so θ=π6\theta = \tfrac{\pi}{6}), then work outward: cot⁡θ\cot\theta, then tan⁡−1(cot⁡θ)\tan^{-1}(\cot\theta).
  • **sin⁡−1(sin⁡x)\sin^{-1}(\sin x):** equals xx only for x∈[−π2,π2]x \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]. Otherwise use sin⁡x=sin⁡(π−x)\sin x = \sin(\pi - x) to bring the angle into range (e.g. sin⁡−1(sin⁡2π3)=π3\sin^{-1}(\sin\tfrac{2\pi}{3}) = \tfrac{\pi}{3}).
  • For cot⁡2(sec⁡−12)+tan⁡2(csc⁡−13)\cot^2(\sec^{-1}2) + \tan^2(\csc^{-1}\sqrt3)-type sums, evaluate each inverse to a standard angle first.

Principal-range reduction

sin⁡−1(sin⁡x)=x  only if x∈[−π2,π2]\sin^{-1}(\sin x) = x \ \text{ only if } x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]

Worked example

Evaluate sin⁡−1 ⁣(sin⁡5π6)\sin^{-1}\!\left(\sin\dfrac{5\pi}{6}\right).
Practice this concept3 quick reps

From the bank · past-year question

Example 1Inverse TrigonometryMODERATE
What is tan⁡−1cot⁡(csc⁡−12)\tan^{-1}\cot(\csc^{-1}2) equal to?

[Q49 · Sep · 2022]

sin⁻¹(sin x) ≠ x outside the principal range

sin⁡−1(sin⁡2π3)\sin^{-1}(\sin\tfrac{2\pi}{3}) is NOT 2π3\tfrac{2\pi}{3} (that is outside [−π2,π2][-\tfrac{\pi}{2},\tfrac{\pi}{2}]). Reduce the inner angle into the principal range first.

Each cancellation uses a DIFFERENT reduction rule

For xx just past the range: sin⁡−1(sin⁡x)=π−x\sin^{-1}(\sin x)=\pi-x, but cos⁡−1(cos⁡x)=2π−x\cos^{-1}(\cos x)=2\pi-x (for x∈[π,2π]x\in[\pi,2\pi]) and tan⁡−1(tan⁡x)=x−π\tan^{-1}(\tan x)=x-\pi. Don't reuse the π−x\pi-x rule for all three — match the reduction to the function's own principal range.
Drill 2 more on inner-to-outer evaluation & sin⁻¹(sin x)

Concept 2 of 3

Double- & Half-Angle Compositions

Intuition

When the expression is tan(2 tan⁻¹x), or a half-angle of an inverse, name the inverse as θ and apply the double- or half-angle formula. The inverse just hands you tan θ (or cos θ); the formula does the rest.

Definition

Set θ=(the inner inverse)\theta = (\text{the inner inverse}), so its argument gives tan⁡θ\tan\theta (or sin⁡θ,cos⁡θ\sin\theta, \cos\theta via a triangle), then apply:

  • Double angle: tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}, sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta.
  • Half angle: tan⁡θ2=1−cos⁡θsin⁡θ=sin⁡θ1+cos⁡θ\tan\tfrac{\theta}{2} = \dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1+\cos\theta}.

Useful for tan⁡(2tan⁡−1x)\tan(2\tan^{-1}x), tan⁡ ⁣(12sec⁡−1t)\tan\!\left(\tfrac12 \sec^{-1}t\right), and 1+sin⁡(2cos⁡−1t)\sqrt{1 + \sin(2\cos^{-1}t)}.

Double-angle tangent

tan⁡(2tan⁡−1x)=2x1−x2\tan(2\tan^{-1}x) = \dfrac{2x}{1 - x^2}

Worked example

Evaluate tan⁡ ⁣(2tan⁡−115)\tan\!\left(2\tan^{-1}\tfrac{1}{5}\right).
Practice this concept2 quick reps

From the bank · past-year question

Example 2Inverse TrigonometryMODERATE
What is tan⁡{2tan⁡−1(13)}\tan\left\{2\tan^{-1}\left(\dfrac{1}{3}\right)\right\} equal to ?

[Q45 · Sep · 2019]

Double-angle tangent has 1 − tan²θ, not 1 + tan²θ

tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \dfrac{2\tan\theta}{1-\tan^2\theta}. The denominator is 1−tan⁡2θ1-\tan^2\theta; using 1+tan⁡2θ1+\tan^2\theta (which is sec⁡2θ\sec^2\theta) is a common confusion that produces the wrong value.
Drill 5 more on double- & half-angle compositions

Concept 3 of 3

Converting Everything to a Tangent

Intuition

A sum of different inverse functions (sin⁻¹ here, cot⁻¹ there) is messy until you rewrite each as a tan⁻¹ using a right triangle. Once every term is a tan⁻¹, the sum/difference formula combines them in one step.

Definition

For each inverse, build the right triangle to read its tangent: e.g. sin⁡−135⇒tan⁡=34\sin^{-1}\tfrac35 \Rightarrow \tan = \tfrac34; cot⁡−132⇒tan⁡=23\cot^{-1}\tfrac32 \Rightarrow \tan = \tfrac23; csc⁡−1414⇒tan⁡=45\csc^{-1}\tfrac{\sqrt{41}}{4} \Rightarrow \tan = \tfrac45. Then combine the resulting tan⁡−1\tan^{-1} terms with the sum/difference formula, and apply the outer function (cot⁡\cot, tan⁡\tan, etc.).

Triangle → tangent

sin⁡−135=tan⁡−134,cot⁡−132=tan⁡−123\sin^{-1}\tfrac{3}{5} = \tan^{-1}\tfrac{3}{4}, \quad \cot^{-1}\tfrac{3}{2} = \tan^{-1}\tfrac{2}{3}

Worked example

Evaluate tan⁡ ⁣(sin⁡−135+cos⁡−11213)\tan\!\left(\sin^{-1}\tfrac{3}{5} + \cos^{-1}\tfrac{12}{13}\right).
Practice this concept2 quick reps

From the bank · past-year question

Example 3Inverse TrigonometryMODERATE
What is the value of cot⁡ ⁣[sin⁡−1 ⁣35+cot⁡−1 ⁣32]\cot\!\left[\sin^{-1}\!\frac{3}{5}+\cot^{-1}\!\frac{3}{2}\right]?

[Q31 · Sep · 2021]

Convert sin⁻¹/cos⁻¹ to tan⁻¹ via the TRIANGLE, not the value

For sin⁡−135\sin^{-1}\tfrac35, the tangent is 34\tfrac34 (opposite 3, adjacent 52−32=4\sqrt{5^2-3^2}=4) — NOT 35\tfrac35. Build the right triangle and read off the missing side before taking the tangent.
Drill 1 more on converting everything to a tangent

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Inner-to-Outer Evaluation & sin⁻¹(sin x)

    Principal-range reduction

    sin⁡−1(sin⁡x)=x  only if x∈[−π2,π2]\sin^{-1}(\sin x) = x \ \text{ only if } x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]
  • Double- & Half-Angle Compositions

    Double-angle tangent

    tan⁡(2tan⁡−1x)=2x1−x2\tan(2\tan^{-1}x) = \dfrac{2x}{1 - x^2}
  • Converting Everything to a Tangent

    Triangle → tangent

    sin⁡−135=tan⁡−134,cot⁡−132=tan⁡−123\sin^{-1}\tfrac{3}{5} = \tan^{-1}\tfrac{3}{4}, \quad \cot^{-1}\tfrac{3}{2} = \tan^{-1}\tfrac{2}{3}

Watch out for (4)

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