The Six Inverse Functions and Their Principal Branches

Trigonometric functions are many-one on their natural domains, so none of them has an inverse until the domain is restricted. Each restriction chosen is the principal value branch, and the whole chapter lives inside these branches:

Function Domain Principal value branch (range)
sin⁡−1x\sin^{-1} x [−1,1][-1, 1] [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
cos⁡−1x\cos^{-1} x [−1,1][-1, 1] [0,π][0, \pi]
tan⁡−1x\tan^{-1} x R\mathbb{R} (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
cot⁡−1x\cot^{-1} x R\mathbb{R} (0,π)(0, \pi)
sec⁡−1x\sec^{-1} x ∣x∣≥1\vert x \vert \geq 1 [0,π]−{π2}[0, \pi] - \left\{\frac{\pi}{2}\right\}
cosec−1 x\mathrm{cosec}^{-1}\, x ∣x∣≥1\vert x \vert \geq 1 [−π2,π2]−{0}\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}

Sine restriction giving the inverse and the six principal branch intervals

The memory pattern: sin⁡−1\sin^{-1}, tan⁡−1\tan^{-1}, cosec−1\mathrm{cosec}^{-1} live in the symmetric interval around 00 (fourth-and-first quadrant); cos⁡−1\cos^{-1}, cot⁡−1\cot^{-1}, sec⁡−1\sec^{-1} live in [0,π][0, \pi] territory (first-and-second quadrant). Also: sin⁡−1x\sin^{-1}x means the inverse function, never 1sin⁡x\frac{1}{\sin x} — the reciprocal is written (sin⁡x)−1(\sin x)^{-1}.

Evaluating principal values

To evaluate cos⁡−1(−12)\cos^{-1}\left(-\frac{1}{2}\right): ask which angle in [0,π][0, \pi] has cosine −12-\frac{1}{2}? — answer 2π3\frac{2\pi}{3}. The branch decides everything: −2π3-\frac{2\pi}{3} and 4π3\frac{4\pi}{3} also have cosine −12-\frac{1}{2}, but they are outside [0,π][0, \pi] and therefore wrong answers.

Negative-argument shortcuts (valid on the whole domain): sin⁡−1(−x)=−sin⁡−1x,tan⁡−1(−x)=−tan⁡−1x,cos⁡−1(−x)=π−cos⁡−1x\sin^{-1}(-x) = -\sin^{-1}x, \qquad \tan^{-1}(-x) = -\tan^{-1}x, \qquad \cos^{-1}(-x) = \pi - \cos^{-1}x Odd behaviour for the symmetric-branch trio; the π\pi-minus flip for the [0,π][0, \pi] trio.

The wrap-around technique

sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta only when θ\theta already lies in the principal branch. Otherwise, first replace θ\theta by the branch angle with the same sine:

  1. sin⁡−1(sin⁡3π5)\sin^{-1}\left(\sin\frac{3\pi}{5}\right): since 3π5>π2\frac{3\pi}{5} > \frac{\pi}{2}, use sin⁡θ=sin⁡(π−θ)\sin\theta = \sin(\pi - \theta): answer π−3π5=2π5\pi - \frac{3\pi}{5} = \frac{2\pi}{5}.
  2. cos⁡−1(cos⁡7π6)\cos^{-1}\left(\cos\frac{7\pi}{6}\right): since 7π6>π\frac{7\pi}{6} > \pi, use cos⁡θ=cos⁡(2π−θ)\cos\theta = \cos(2\pi - \theta): answer 2π−7π6=5π62\pi - \frac{7\pi}{6} = \frac{5\pi}{6}.
  3. tan⁡−1(tan⁡3π4)\tan^{-1}\left(\tan\frac{3\pi}{4}\right): use tan⁡θ=tan⁡(θ−π)\tan\theta = \tan(\theta - \pi): answer 3π4−π=−π4\frac{3\pi}{4} - \pi = -\frac{\pi}{4}.

Answering 3π5\frac{3\pi}{5}, 7π6\frac{7\pi}{6} or 3π4\frac{3\pi}{4} to these is the classic error of the chapter — always check the branch before cancelling.

Worked Examples — Principal Values and Wrap-Arounds

Example 1 — Direct principal values

Evaluate (i) sin⁡−1(12)\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right), (ii) cos⁡−1(−12)\cos^{-1}\left(-\dfrac{1}{2}\right), (iii) tan⁡−1(−1)\tan^{-1}(-1).

Step 1 — (i): which angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] has sine 12\frac{1}{\sqrt 2}? — π4\frac{\pi}{4}.

Step 2 — (ii): which angle in [0,π][0, \pi] has cosine −12-\frac{1}{2}? — π−π3=2π3\pi - \frac{\pi}{3} = \frac{2\pi}{3}.

Step 3 — (iii): odd function: tan⁡−1(−1)=−tan⁡−1(1)=−π4\tan^{-1}(-1) = -\tan^{-1}(1) = -\frac{\pi}{4}.

Answer: π4\frac{\pi}{4}, 2π3\frac{2\pi}{3}, −π4-\frac{\pi}{4} — each pinned by its own branch.

Example 2 — A negative argument in the reciprocal family

Find the principal value of cosec−1(−2)\mathrm{cosec}^{-1}(-\sqrt{2}).

Step 1 — odd function: cosec−1(−x)=−cosec−1x\mathrm{cosec}^{-1}(-x) = -\mathrm{cosec}^{-1}x.

Step 2 — the positive value: cosec−1(2)\mathrm{cosec}^{-1}(\sqrt 2) asks for the branch angle with cosecant 2\sqrt 2, i.e. sine 12\frac{1}{\sqrt 2}: that is π4\frac{\pi}{4}.

Answer: −π4-\dfrac{\pi}{4} — inside the branch [−π2,π2]−{0}\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}. ✓

Example 3 — Wrap-around with sine

Evaluate sin⁡−1(sin⁡3π5)\sin^{-1}\left(\sin\dfrac{3\pi}{5}\right).

Step 1 — check the branch: 3π5\frac{3\pi}{5} exceeds π2\frac{\pi}{2} — direct cancellation is illegal.

Step 2 — same sine, right branch: sin⁡3π5=sin⁡(π−3π5)=sin⁡2π5\sin\frac{3\pi}{5} = \sin\left(\pi - \frac{3\pi}{5}\right) = \sin\frac{2\pi}{5}, and 2π5\frac{2\pi}{5} is in the branch.

Answer: 2π5\dfrac{2\pi}{5} — never 3π5\frac{3\pi}{5}.

Example 4 — Wrap-around with cosine and tangent

Evaluate (i) cos⁡−1(cos⁡7π6)\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right), (ii) tan⁡−1(tan⁡3π4)\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right).

Step 1 — (i): 7π6∉[0,π]\frac{7\pi}{6} \notin [0, \pi]; same cosine at 2π−7π6=5π62\pi - \frac{7\pi}{6} = \frac{5\pi}{6}, which is in the branch.

Step 2 — (ii): 3π4∉(−π2,π2)\frac{3\pi}{4} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right); tangent has period π\pi, so subtract: 3π4−π=−π4\frac{3\pi}{4} - \pi = -\frac{\pi}{4}.

Answer: 5π6\frac{5\pi}{6} and −π4-\frac{\pi}{4} — the correction rule depends on the function (π−θ\pi - \theta for sine, 2π−θ2\pi - \theta for cosine, θ−π\theta - \pi for tangent).

Example 5 — A three-term principal-value sum

Evaluate tan⁡−1(1)+cos⁡−1(−12)+sin⁡−1(−12)\tan^{-1}(1) + \cos^{-1}\left(-\dfrac{1}{2}\right) + \sin^{-1}\left(-\dfrac{1}{2}\right).

Step 1 — each term by its branch: tan⁡−1(1)=π4\tan^{-1}(1) = \frac{\pi}{4}; cos⁡−1(−12)=2π3\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}; sin⁡−1(−12)=−π6\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}.

Step 2 — add with a common denominator: 3π12+8π12−2π12=9π12\frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12}.

Answer: 3π4\dfrac{3\pi}{4} — note how the negative-argument rules do all the sign work.

Properties — Cancellation, Complements and Simplest Forms

The cancellation laws, with their fine print

sin⁡(sin⁡−1x)=x    (−1≤x≤1),sin⁡−1(sin⁡θ)=θ    (−π2≤θ≤π2)\sin(\sin^{-1}x) = x \;\; (-1 \leq x \leq 1), \qquad \sin^{-1}(\sin\theta) = \theta \;\; \left(-\tfrac{\pi}{2} \leq \theta \leq \tfrac{\pi}{2}\right)

The first law is unconditional on the domain; the second demands the branch — that asymmetry is the source of every wrap-around question. The same pattern holds for all six functions.

The complementary pairs

sin⁡−1x+cos⁡−1x=π2,tan⁡−1x+cot⁡−1x=π2,sec⁡−1x+cosec−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \qquad \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \qquad \sec^{-1}x + \mathrm{cosec}^{-1}x = \frac{\pi}{2}

(each on the appropriate domain). These convert mixed sums instantly — any expression containing sin⁡−1x+cos⁡−1x\sin^{-1}x + \cos^{-1}x is hiding a constant π2\frac{\pi}{2}.

Reciprocal conversions

sin⁡−11x=cosec−1x,cos⁡−11x=sec⁡−1x,(∣x∣≥1)\sin^{-1}\frac{1}{x} = \mathrm{cosec}^{-1}x, \qquad \cos^{-1}\frac{1}{x} = \sec^{-1}x, \qquad (\vert x \vert \geq 1)

so sec⁡−1(2)=cos⁡−112=π3\sec^{-1}(2) = \cos^{-1}\frac{1}{2} = \frac{\pi}{3} — always convert the reciprocal family to sine/cosine/tangent before computing.

Simplest forms — the substitution dictionary

"Write in simplest form" questions all yield to one idea: substitute a trig function for xx so the expression collapses to a single angle.

Expression contains Substitute Typical collapse
a2−x2\sqrt{a^2 - x^2} x=asin⁡θx = a\sin\theta tan⁡−1xa2−x2=sin⁡−1xa\tan^{-1}\frac{x}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}
1−cos⁡x\sqrt{1 - \cos x}, 1+cos⁡x\sqrt{1 + \cos x} half-angle tan⁡−11−cos⁡x1+cos⁡x=x2\tan^{-1}\frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}} = \frac{x}{2}
cos⁡x1±sin⁡x\frac{\cos x}{1 \pm \sin x} half-angle split tan⁡−1cos⁡x1−sin⁡x=π4+x2\tan^{-1}\frac{\cos x}{1 - \sin x} = \frac{\pi}{4} + \frac{x}{2}
a2+x2\sqrt{a^2 + x^2} x=atan⁡θx = a\tan\theta radical becomes asec⁡θa\sec\theta

Presentation for full marks: name the substitution, show the collapse to tan⁡−1(tan⁡(⋅))\tan^{-1}(\tan(\cdot)) or sin⁡−1(sin⁡(⋅))\sin^{-1}(\sin(\cdot)), state the interval that keeps the angle inside the principal branch, and only then cancel. The interval sentence is a genuine mark on the board scheme, not decoration.

Evaluating trig-of-inverse-trig: for tan⁡(sin⁡−135)\tan\left(\sin^{-1}\frac{3}{5}\right), set θ=sin⁡−135\theta = \sin^{-1}\frac{3}{5}, draw the 33-44-55 right triangle (opposite 33, hypotenuse 55, adjacent 44), and read off tan⁡θ=34\tan\theta = \frac{3}{4}. The triangle method turns every such evaluation into arithmetic.

Worked Examples — Properties and Simplest Forms

Example 6 — Complementary pair in action

Evaluate sin⁡−1(27)+cos⁡−1(27)\sin^{-1}\left(\dfrac{2}{7}\right) + \cos^{-1}\left(\dfrac{2}{7}\right).

Step 1 — recognise the pair: the two arguments are identical and lie in [−1,1][-1, 1].

Answer: π2\dfrac{\pi}{2}, with no computation — the complementary identity does not care what the common argument is.

Example 7 — Reciprocal conversion

Find the principal value of sec⁡−1(2)\sec^{-1}(2).

Step 1 — convert: sec⁡−1(2)=cos⁡−112\sec^{-1}(2) = \cos^{-1}\frac{1}{2}.

Step 2 — evaluate in [0,π][0, \pi]: cos⁡−112=π3\cos^{-1}\frac{1}{2} = \frac{\pi}{3}.

Answer: π3\dfrac{\pi}{3} — the reciprocal family always routes through its partner function.

Example 8 — The triangle method

Evaluate tan⁡(sin⁡−135)\tan\left(\sin^{-1}\dfrac{3}{5}\right) and cos⁡(sin⁡−135)\cos\left(\sin^{-1}\dfrac{3}{5}\right).

Step 1 — set the angle: θ=sin⁡−135\theta = \sin^{-1}\frac{3}{5}, so sin⁡θ=35\sin\theta = \frac{3}{5} with θ\theta in the first quadrant.

Step 2 — build the triangle: opposite 33, hypotenuse 55, so adjacent =25−9=4= \sqrt{25 - 9} = 4.

Step 3 — read off: tan⁡θ=34\tan\theta = \frac{3}{4} and cos⁡θ=45\cos\theta = \frac{4}{5}.

Answer: 34\frac{3}{4} and 45\frac{4}{5} — one triangle answers every ratio question about the same angle.

Example 9 — Simplest form with half-angles

Write tan⁡−1(1−cos⁡x1+cos⁡x)\tan^{-1}\left(\dfrac{\sqrt{1 - \cos x}}{\sqrt{1 + \cos x}}\right), 0<x<π0 < x < \pi, in simplest form.

Step 1 — half-angle identities: 1−cos⁡x=2sin⁡2x21 - \cos x = 2\sin^2\frac{x}{2} and 1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\frac{x}{2}.

Step 2 — collapse the quotient: on 0<x<π0 < x < \pi both half-angle values are positive, so the roots strip cleanly: tan⁡−1(sin⁡x2cos⁡x2)=tan⁡−1(tan⁡x2)\tan^{-1}\left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\right) = \tan^{-1}\left(\tan\frac{x}{2}\right)

Step 3 — cancel inside the branch: 0<x2<π20 < \frac{x}{2} < \frac{\pi}{2}, safely inside (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

Answer: x2\dfrac{x}{2} — with the interval check written, which is where the last mark lives.

Example 10 — The quarter-turn form

Write tan⁡−1(cos⁡x1−sin⁡x)\tan^{-1}\left(\dfrac{\cos x}{1 - \sin x}\right), −π2<x<π2-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}, in simplest form.

Step 1 — half-angle split: cos⁡x=cos⁡2x2−sin⁡2x2\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} and 1−sin⁡x=(cos⁡x2−sin⁡x2)21 - \sin x = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2.

Step 2 — factor and cancel: (cos⁡x2+sin⁡x2)(cos⁡x2−sin⁡x2)(cos⁡x2−sin⁡x2)2=cos⁡x2+sin⁡x2cos⁡x2−sin⁡x2=1+tan⁡x21−tan⁡x2=tan⁡(π4+x2)\frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2} = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}} = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)

Step 3 — cancel inside the branch: for the given interval, π4+x2\frac{\pi}{4} + \frac{x}{2} stays in (0,π2)\left(0, \frac{\pi}{2}\right).

Answer: π4+x2\dfrac{\pi}{4} + \dfrac{x}{2}.

Example 11 — Substitution with a radical

Write tan⁡−1(xa2−x2)\tan^{-1}\left(\dfrac{x}{\sqrt{a^2 - x^2}}\right), ∣x∣<a\vert x \vert < a, in simplest form.

Step 1 — substitute x=asin⁡θx = a\sin\theta: then a2−x2=acos⁡θ\sqrt{a^2 - x^2} = a\cos\theta and the fraction is tan⁡θ\tan\theta.

Step 2 — collapse: tan⁡−1(tan⁡θ)=θ\tan^{-1}(\tan\theta) = \theta, valid since θ=sin⁡−1xa\theta = \sin^{-1}\frac{x}{a} lies in the tangent branch.

Answer: sin⁡−1xa\sin^{-1}\dfrac{x}{a} — the a2−x2\sqrt{a^2 - x^2} shape always calls the sine substitution.

Example 12 — Solving a simple inverse-trig equation

Find xx if tan⁡−1x+2cot⁡−1x=2π3\tan^{-1}x + 2\cot^{-1}x = \dfrac{2\pi}{3}.

Step 1 — trade cot⁡−1\cot^{-1} for tan⁡−1\tan^{-1}: write 2cot⁡−1x=2(π2−tan⁡−1x)=π−2tan⁡−1x2\cot^{-1}x = 2\left(\frac{\pi}{2} - \tan^{-1}x\right) = \pi - 2\tan^{-1}x.

Step 2 — substitute and solve: tan⁡−1x+π−2tan⁡−1x=2π3\tan^{-1}x + \pi - 2\tan^{-1}x = \frac{2\pi}{3} gives tan⁡−1x=π−2π3=π3\tan^{-1}x = \pi - \frac{2\pi}{3} = \frac{\pi}{3}.

Step 3 — invert: x=tan⁡π3=3x = \tan\frac{\pi}{3} = \sqrt{3}.

Answer: x=3x = \sqrt{3} — check: tan⁡−13+2cot⁡−13=π3+2⋅π6=2π3\tan^{-1}\sqrt 3 + 2\cot^{-1}\sqrt 3 = \frac{\pi}{3} + 2\cdot\frac{\pi}{6} = \frac{2\pi}{3}. ✓