Inverse trig identities

Ask a student to evaluate sin⁡−135+sin⁡−1817\sin^{-1}\frac35 + \sin^{-1}\frac{8}{17} and the natural instinct is to reach for a calculator. ISC Class 12 wants something better: a toolkit of algebraic identities that turns a sum, difference, or multiple of inverse trig terms into a single clean inverse trig expression, without ever leaving exact fractions. This is the identity layer built directly on top of the definitions, domains, and principal-value branches already established for sin⁡−1,cos⁡−1,tan⁡−1\sin^{-1}, \cos^{-1}, \tan^{-1} and their reciprocals. The payoff is speed and certainty on proof-style questions: collapse tan⁡−112+tan⁡−113\tan^{-1}\frac12+\tan^{-1}\frac13 into tan⁡−1(1)\tan^{-1}(1) and the answer π4\frac{\pi}{4} falls out in one line, provided the validity condition guarding the formula is checked first. That check — and knowing when a hidden π\pi-correction is needed — is what separates a confident answer from a silently wrong one. This continues directly from the definitions and principal-value branches of the inverse trig functions; solving general trigonometric equations and differentiating inverse trig expressions stay out of scope here, since this unit is purely about the algebra of the identities themselves.

The core insight

A sum, difference, or multiple of inverse trig terms of the same or related arguments can be rewritten as a single inverse trig expression using conversion, sum/difference, and double/triple-angle identities, provided the accompanying validity condition on the arguments is satisfied.

The core idea

Three families of identity do all the work in this unit: identities that convert between inverse trig functions of the same argument, identities that combine two different arguments into one via a sum or difference, and identities that collapse a multiple of one inverse trig term (double or triple) into a single term. Each family reduces an expression to something a calculator-free exam can check exactly, and each comes with a validity condition that must be verified before the formula is trusted.

1. Conversion between inverse trig functions of the same argument. For suitable xx (typically x>0x>0 unless stated otherwise), the three functions sin⁡−1x\sin^{-1}x, cos⁡−1x\cos^{-1}x, tan⁡−1x\tan^{-1}x of the same argument are all expressible in terms of one another, since they name three different sides of one right triangle:

sin⁡−1x=cos⁡−11−x2=tan⁡−1x1−x2,sin⁡−1x+cos⁡−1x=π2\sin^{-1}x = \cos^{-1}\sqrt{1-x^2} = \tan^{-1}\frac{x}{\sqrt{1-x^2}}, \qquad \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}

where: xx is the shared argument, restricted so every root stays real and every angle stays in its principal branch. The analogous complementary and reciprocal relations hold for cos⁡−1,tan⁡−1,cot⁡−1,sec⁡−1,csc⁡−1\cos^{-1}, \tan^{-1}, \cot^{-1}, \sec^{-1}, \csc^{-1}, and each one is read straight off a right triangle with opposite side xx, hypotenuse 11, and adjacent side 1−x2\sqrt{1-x^2}.

Figure
A right triangle with opposite side x hypotenuse 1 and adjacent side square root of 1 minus x squared showing how sine cosine and tangent inverse of the same argument are read off one triangle
One triangle, three inverse trig readings x (opposite) 1 (hypotenuse) square root of 1 minus x squared (adjacent) theta theta = sin inverse x = cos inverse of the adjacent ratio = tan inverse of opposite over adjacent

2. Sum and difference formulae. These combine two different arguments into one:

sin⁡−1x±sin⁡−1y=sin⁡−1 ⁣(x1−y2±y1−x2)\sin^{-1}x\pm\sin^{-1}y=\sin^{-1}\!\left(x\sqrt{1-y^2}\pm y\sqrt{1-x^2}\right)
cos⁡−1x±cos⁡−1y=cos⁡−1 ⁣(xy∓1−x21−y2)\cos^{-1}x\pm\cos^{-1}y=\cos^{-1}\!\left(xy\mp\sqrt{1-x^2}\sqrt{1-y^2}\right)
tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy (xy<1),tan⁡−1x−tan⁡−1y=tan⁡−1x−y1+xy (xy>−1)\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\ (xy<1), \qquad \tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}\ (xy>-1)

where: x,yx,y are the two inverse-trig arguments, and the bracketed conditions xy<1xy<1 or xy>−1xy>-1 are validity conditions guaranteeing the combined angle lands back inside the principal branch (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right); outside that range the raw formula is off by exactly π\pi.

3. Double and triple-angle formulae, obtained by setting y=xy=x in the sum formulae:

2sin⁡−1x=sin⁡−1 ⁣(2x1−x2),2cos⁡−1x=cos⁡−1(2x2−1),2tan⁡−1x=tan⁡−12x1−x22\sin^{-1}x = \sin^{-1}\!\left(2x\sqrt{1-x^2}\right), \qquad 2\cos^{-1}x = \cos^{-1}(2x^2-1), \qquad 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}
3tan⁡−1x=tan⁡−13x−x31−3x23\tan^{-1}x = \tan^{-1}\frac{3x-x^3}{1-3x^2}

Working left to right collapses a sum or multiple into one term — the fastest route to proving an equation. Working right to left, expanding one term into a sum, is equally useful when an expression needs splitting before further simplification. Because the double-angle formulae are just the sum formulae with y=xy=x, memorising the sum formulae alone is enough to reconstruct every double-angle result under exam pressure, which is worth knowing when memory of the full identity list is shaky.

Derivation AWhy the tan-inverse sum needs xy less than 1 and how the collapse works

The goal of this derivation is to show, entirely from the tangent addition formula a student already knows and the definition of the principal branch of tan⁡−1\tan^{-1}, both why tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy} holds whenever xy<1xy<1, and why that condition is not optional decoration but the exact statement that keeps the combined angle inside (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) so the formula can be inverted safely.

Step 1. Name the two angles. Let α=tan⁡−1x\alpha=\tan^{-1}x and β=tan⁡−1y\beta=\tan^{-1}y, so by definition tan⁡α=x\tan\alpha=x, tan⁡β=y\tan\beta=y, with α,β∈(−π2,π2)\alpha,\beta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) — the principal branch of tan⁡−1\tan^{-1}.

Step 2. Apply the ordinary tangent addition formula, which holds for any two angles whose tangents are defined: tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=x+y1−xy\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\dfrac{x+y}{1-xy}.

Step 3. Decide whether α+β\alpha+\beta itself lies in the principal branch. Since each of α,β\alpha,\beta individually lies in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right), their sum lies in (−π,π)(-\pi,\pi) — a wider range than tan⁡−1\tan^{-1}'s own principal branch. Only when α+β\alpha+\beta additionally sits inside (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) can we take tan⁡−1\tan^{-1} of both sides of Step 2 and recover α+β=tan⁡−1x+y1−xy\alpha+\beta=\tan^{-1}\dfrac{x+y}{1-xy} directly.

Step 4. Translate that geometric requirement into the algebraic condition on xx and yy. It can be shown that α+β\alpha+\beta stays inside (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) precisely when xy<1xy<1; when xy>1xy>1 instead, α+β\alpha+\beta has overflowed the branch and the true sum equals tan⁡−1x+y1−xy+π\tan^{-1}\frac{x+y}{1-xy}+\pi (if x,y>0x,y>0) or minus π\pi (if x,y<0x,y<0), since the raw right-hand side then names the wrong representative of the same tangent value. Verified: with x=12,y=13x=\tfrac12,y=\tfrac13, xy=16<1xy=\tfrac16<1 satisfies the condition and the formula gives tan⁡−1(1)=π4\tan^{-1}(1)=\tfrac{\pi}{4}, matching the worked example below and the source article's proof exactly.

Worked example

Prove that sin⁡−135+sin⁡−1817=sin⁡−17785\sin^{-1}\dfrac{3}{5} + \sin^{-1}\dfrac{8}{17} = \sin^{-1}\dfrac{77}{85}.

Step 1. Set x=35x=\frac35, y=817y=\frac{8}{17} and write down the sum formula to be used: sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2)\sin^{-1}x+\sin^{-1}y = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right).

Step 2. Compute the two square roots needed. 1−x2=1−925=16251-x^2 = 1-\frac{9}{25} = \frac{16}{25}, so 1−x2=45\sqrt{1-x^2}=\frac45. And 1−y2=1−64289=2252891-y^2 = 1-\frac{64}{289} = \frac{225}{289}, so 1−y2=1517\sqrt{1-y^2}=\frac{15}{17}.

Step 3. Substitute into the formula. x1−y2+y1−x2=35⋅1517+817⋅45=4585+3285=7785x\sqrt{1-y^2} + y\sqrt{1-x^2} = \frac35\cdot\frac{15}{17} + \frac{8}{17}\cdot\frac45 = \frac{45}{85} + \frac{32}{85} = \frac{77}{85}.

Step 4. Check the branch before declaring the answer final. Both 35\frac35 and 817\frac{8}{17} are positive and less than 11, so both angles lie in (0,π2)\left(0,\frac{\pi}{2}\right); the combined sine 7785\frac{77}{85} is still less than 11 and the sum of two acute-ish angles here stays short of π2\frac{\pi}{2}, so no supplementary correction is needed. Therefore sin⁡−135+sin⁡−1817=sin⁡−17785\sin^{-1}\frac35+\sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}.

Competitive stretch

A natural follow-up: what is 2tan⁡−1132\tan^{-1}\frac13? Using the double-angle formula with x=13x=\frac13: 2tan⁡−113=tan⁡−12⋅131−19=tan⁡−12389=tan⁡−1(23⋅98)=tan⁡−1342\tan^{-1}\frac13 = \tan^{-1}\dfrac{2\cdot\frac13}{1-\frac19} = \tan^{-1}\dfrac{\frac23}{\frac89} = \tan^{-1}\left(\frac23\cdot\frac98\right) = \tan^{-1}\frac34. Since 13⋅13=19<1\frac13\cdot\frac13=\frac19<1, no correction is needed, so 2tan⁡−113=tan⁡−1342\tan^{-1}\frac13=\tan^{-1}\frac34 — a compact cross-check that the double-angle formula is just the sum formula with y=xy=x.

Common traps

The identities above are algebraically simple, but nearly every mark lost on this topic comes from skipping the branch check, not from a wrong formula. Watch for these four patterns.

  • Applying the tan⁡−1x+tan⁡−1y\tan^{-1}x+\tan^{-1}y formula without checking xy<1xy<1 (or xy>−1xy>-1 for the difference) — outside that range the raw formula's angle is wrong by exactly π\pi.
  • Substituting into the sin⁡−1\sin^{-1} or cos⁡−1\cos^{-1} sum formulae without checking that both individual angles and the combined result stay in the expected branch, which can silently need a π−(⋅)\pi-(\cdot) correction.
  • Flipping the sign inside cos⁡−1x±cos⁡−1y=cos⁡−1(xy∓1−x21−y2)\cos^{-1}x\pm\cos^{-1}y=\cos^{-1}\left(xy\mp\sqrt{1-x^2}\sqrt{1-y^2}\right) — the inner sign is opposite to the sign on the left, easy to reverse under exam pressure.
  • Arithmetic slips simplifying 1−x2\sqrt{1-x^2} for a fractional xx — always reduce 1−x21-x^2 to a single fraction before taking the square root.

Ask the tutor: If a sum/difference or multiple-angle inverse-trig proof is not matching the expected result, ask the AI tutor to walk through the validity-condition check and pinpoint exactly where the branch may have overflowed.

How do I know whether a sum/difference identity needs a pi correction?

Check where each individual angle sits first. For the tan⁡−1\tan^{-1} sum, work out xyxy: if xy<1xy<1 the raw formula's angle already lies in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) and no correction is needed; if xy>1xy>1 with x,y>0x,y>0, add π\pi. For sin⁡−1\sin^{-1} or cos⁡−1\cos^{-1} sums, check that both arguments give acute or otherwise compatible principal angles — if the combined angle would exceed the branch, subtract from π\pi instead of trusting the raw output.

Why do the double and triple-angle formulae look so different from the ordinary sin 2 theta and tan 2 theta formulae?

They are not actually different — they are the same multiple-angle trig identities, just read backwards through an inverse function. Setting θ=tan⁡−1x\theta=\tan^{-1}x in the ordinary tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta} and simplifying gives exactly 2tan⁡−1x=tan⁡−12x1−x22\tan^{-1}x=\tan^{-1}\dfrac{2x}{1-x^2}. The sum formula with y=xy=x produces the identical result, which is a useful self-check when deriving these from memory.

Do I need to prove the validity condition every time, or can I just quote it?

On a proof-type question, at minimum state the condition (like xy<1xy<1) and confirm the given numbers satisfy it — that one line is usually worth marks and protects you from a silent branch error. You do not need to re-derive why the condition arises from scratch every time; quoting it correctly and checking it against the given values is standard and expected exam practice.

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