Inverse Trigonometric Functions

Study Sheet

Inverse Trigonometric Functions

Definitions, exact values, compositions, graphs & equations

Definitions: Domain & Range

An inverse trig function undoes a trig function. Because sin⁡,cos⁡,tan⁡\sin,\cos,\tan are not one-to-one, we first restrict their domains so an inverse can exist. Those restrictions become the ranges of the inverse functions.

Concept
The three inverse trig functions

y=arcsin⁡xy=\arcsin x means “yy is the angle whose sine is xx”: sin⁡y=x\sin y = x. The same idea defines arccos⁡\arccos and arctan⁡\arctan. Read sin⁡−1x=arcsin⁡x\sin^{-1}x=\arcsin x (the −1-1 is not an exponent).

The range tells you which quadrant the output angle lives in: arcsin⁡\arcsin and arctan⁡\arctan give angles in QI or QIV (right half), while arccos⁡\arccos gives angles in QI or QII (upper half).

Tip

Sign memory aid. arccos⁡\arccos always returns an angle in [0,π][0,\pi], so its output is never negative. arcsin⁡\arcsin and arctan⁡\arctan return negative angles for negative inputs.

Evaluating Exactly and with a Calculator

For special values, ask “what angle in the correct range has this sine/cosine/tangent?”

Example
Exact value with a negative input

Evaluate arccos⁡ ⁣(−22)\arccos\!\left(-\dfrac{\sqrt{2}}{2}\right).

Need angle y with cos⁡y=−22 and y∈[0,π].Reference angle: π4. Cosine is negative in QII.y=π−π4=3π4.\begin{aligned} &\text{Need angle } y \text{ with } \cos y=-\tfrac{\sqrt2}{2} \text{ and } y\in[0,\pi].\\ &\text{Reference angle: } \tfrac{\pi}{4}. \text{ Cosine is negative in QII.}\\ &y=\pi-\tfrac{\pi}{4}=\boxed{\tfrac{3\pi}{4}}. \end{aligned}

Check: 3π4∈[0,π]\tfrac{3\pi}{4}\in[0,\pi], and cos⁡3π4=−22\cos\tfrac{3\pi}{4}=-\tfrac{\sqrt2}{2}. ✓

Example
Exact value for arcsin of a negative

Evaluate arcsin⁡ ⁣(−12)\arcsin\!\left(-\dfrac12\right). Need yy with sin⁡y=−12\sin y=-\tfrac12 and y∈[−π2,π2]y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]. Reference angle π6\tfrac{\pi}{6}; sine negative means QIV (a negative angle): arcsin⁡ ⁣(−12)=−π6.\arcsin\!\left(-\tfrac12\right)=-\dfrac{\pi}{6}.

Tip

Calculator. Put the calculator in radian mode. Use SIN−1^{-1}, COS−1^{-1}, TAN−1^{-1}. Example: arctan⁡(4)≈1.326\arctan(4)\approx 1.326, arccos⁡(−0.72)≈2.375\arccos(-0.72)\approx 2.375. The calculator only ever reports the one value inside the range above.

Compositions with a Reference Triangle

To turn sin⁡(arccos⁡x)\sin(\arccos x), tan⁡(arcsin⁡x)\tan(\arcsin x), etc. into an algebraic expression: let θ\theta be the inner inverse, draw a right triangle in which θ\theta has the required ratio, fill in the missing side with the Pythagorean theorem, then read off the outer function.

An angle in standard position.

Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2
Example
Numeric composition

Evaluate cos⁡ ⁣(arcsin⁡35)\cos\!\left(\arcsin\dfrac{3}{5}\right). Let θ=arcsin⁡35\theta=\arcsin\tfrac35, so sin⁡θ=35=opphyp\sin\theta=\tfrac{3}{5}=\tfrac{\text{opp}}{\text{hyp}} with θ\theta in QI.

Missing leg: 52−32=4\sqrt{5^2-3^2}=4. Then cos⁡θ=adjhyp=45\cos\theta=\dfrac{\text{adj}}{\text{hyp}}=\boxed{\dfrac{4}{5}}.

Example
Algebraic composition

Write tan⁡(arccos⁡x)\tan(\arccos x) as an algebraic expression for 0<x≤10<x\le 1. Let θ=arccos⁡x\theta=\arccos x, so cos⁡θ=x1=adjhyp\cos\theta=\dfrac{x}{1}=\dfrac{\text{adj}}{\text{hyp}}. Take adj =x=x, hyp =1=1, so opp =1−x2=\sqrt{1-x^2}.

tan⁡(arccos⁡x)=oppadj=1−x2x.\tan(\arccos x)=\frac{\text{opp}}{\text{adj}}=\frac{\sqrt{1-x^{2}}}{x}.

Likewise sin⁡(arccos⁡x)=1−x2\sin(\arccos x)=\sqrt{1-x^{2}} and tan⁡(arcsin⁡x)=x1−x2\tan(\arcsin x)=\dfrac{x}{\sqrt{1-x^{2}}}.

Tip

Tip. For an inner arctan⁡x2\arctan\frac{x}{2}, put opp =x=x, adj =2=2, so hyp =x2+4=\sqrt{x^2+4}. The number under the square root is always (first leg)2+(second leg)2(\text{first leg})^2+(\text{second leg})^2 or 1−x21-x^2.

Compositions Where the Range Restriction Matters

arcsin⁡(sin⁡θ)=θ\arcsin(\sin\theta)=\theta only when θ\theta is already inside the inverse's range. Otherwise you must first evaluate the inside, then take the inverse --- the answer is the angle in range that has the same value.

Opposite, adjacent and hypotenuse are named from the angle.

Example
The classic trap

Evaluate arcsin⁡ ⁣(sin⁡5π6)\arcsin\!\left(\sin\dfrac{5\pi}{6}\right). Tempting (wrong) answer: 5π6\tfrac{5\pi}{6}. But 5π6∉[−π2,π2]\tfrac{5\pi}{6}\notin\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right], so the functions do not simply cancel.

sin⁡5π6=12,arcsin⁡ ⁣(12)=π6∈[−π2,π2].\begin{aligned} \sin\tfrac{5\pi}{6}&=\tfrac12,\\ \arcsin\!\left(\tfrac12\right)&=\tfrac{\pi}{6}\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]. \end{aligned}

So arcsin⁡ ⁣(sin⁡5π6)=π6\arcsin\!\left(\sin\tfrac{5\pi}{6}\right)=\boxed{\dfrac{\pi}{6}}.

Example
Cosine version

Evaluate arccos⁡ ⁣(cos⁡7π6)\arccos\!\left(\cos\dfrac{7\pi}{6}\right). 7π6∉[0,π]\tfrac{7\pi}{6}\notin[0,\pi]. Compute inside first: cos⁡7π6=−32\cos\tfrac{7\pi}{6}=-\tfrac{\sqrt3}{2}. Then arccos⁡ ⁣(−32)=5π6\arccos\!\left(-\tfrac{\sqrt3}{2}\right)=\dfrac{5\pi}{6} (the angle in [0,π][0,\pi]).

Tip

Cancellation rules.  arcsin⁡(sin⁡θ)=θ\ \arcsin(\sin\theta)=\theta only if θ∈[−π2,π2]\theta\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right];  arccos⁡(cos⁡θ)=θ\ \arccos(\cos\theta)=\theta only if θ∈[0,π]\theta\in[0,\pi];  arctan⁡(tan⁡θ)=θ\ \arctan(\tan\theta)=\theta only if θ∈(−π2,π2)\theta\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right). In contrast, sin⁡(arcsin⁡x)=x\sin(\arcsin x)=x always holds for x∈[−1,1]x\in[-1,1].

Graphs of the Inverse Functions

Each inverse graph is the reflection of the restricted trig graph across the line y=xy=x. Domain and range trade places.

 

Tip

Read the graphs. arcsin⁡\arcsin and arccos⁡\arccos have endpoints at x=±1x=\pm1 (domain [−1,1][-1,1]). arctan⁡\arctan is defined for all xx and flattens toward the horizontal asymptotes y=±π2y=\pm\frac{\pi}{2}; it never reaches them.

Solving Simple Trig Equations

Inverse functions give one solution; use symmetry (reference angle + quadrants) or add 2πk2\pi k to get the rest.

Example
Two solutions on one revolution

Solve cos⁡x=−12\cos x=-\dfrac12 on [0,2π)[0,2\pi). Reference angle: arccos⁡12=π3\arccos\tfrac12=\tfrac{\pi}{3}. Cosine is negative in QII and QIII:

x=π−π3=2π3andx=π+π3=4π3.x=\pi-\tfrac{\pi}{3}=\tfrac{2\pi}{3} \quad\text{and}\quad x=\pi+\tfrac{\pi}{3}=\tfrac{4\pi}{3}.
Tip

Solving recipe. (1) Isolate the trig function. (2) Take the inverse to get the reference angle. (3) Use the sign to decide the quadrants. (4) List every solution in the requested interval; for “all solutions” add +2πk+2\pi k (sine/cosine) or +πk+\pi k (tangent).

Going Deeper: Advanced Inverse Trig

This section pushes past single evaluations into identities that combine inverse functions, quadrant traps hidden inside the arctangent sum formula, and the domain bookkeeping needed when inverses are composed.

The complementary identity

Opposite, adjacent and hypotenuse are named from the angle.

Concept
arcsin⁡x+arccos⁡x=π2\arcsin x+\arccos x=\dfrac{\pi}{2}

For every x∈[−1,1]x\in[-1,1],

arcsin⁡x+arccos⁡x=π2.\arcsin x+\arccos x=\frac{\pi}{2}.

Why: let α=arcsin⁡x\alpha=\arcsin x, so sin⁡α=x\sin\alpha=x with α∈[−π2,π2]\alpha\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]. Then cos⁡ ⁣(π2−α)=sin⁡α=x\cos\!\left(\tfrac{\pi}{2}-\alpha\right)=\sin\alpha=x, and π2−α∈[0,π]\tfrac{\pi}{2}-\alpha\in[0,\pi] --- exactly the range of arccos⁡\arccos. So π2−α=arccos⁡x\tfrac{\pi}{2}-\alpha=\arccos x. The identity is just “sine and cosine of complementary angles agree.”

Tip

Use it to dodge triangles. Any sum arcsin⁡x+arccos⁡x\arcsin x+\arccos x collapses to π2\tfrac{\pi}{2} instantly. For example arccos⁡x=π2−arcsin⁡x\arccos x=\tfrac{\pi}{2}-\arcsin x, so sin⁡(arccos⁡x)=sin⁡ ⁣(π2−arcsin⁡x)=cos⁡(arcsin⁡x)=1−x2\sin(\arccos x)=\sin\!\left(\tfrac{\pi}{2}-\arcsin x\right)=\cos(\arcsin x)=\sqrt{1-x^2}, matching the triangle method with no drawing.

Algebraic compositions with double angles

Example
Simplify sin⁡(2arctan⁡x)\sin(2\arctan x)

Let θ=arctan⁡x\theta=\arctan x, so tan⁡θ=x1=oppadj\tan\theta=\dfrac{x}{1}=\dfrac{\text{opp}}{\text{adj}} with θ∈(−π2,π2)\theta\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right). Build the triangle: opp =x=x, adj =1=1, hyp =1+x2=\sqrt{1+x^2}.

Apply the double-angle formula sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta=2\sin\theta\cos\theta:

sin⁡(2arctan⁡x)=2⋅x1+x2⋅11+x2=2x1+x2.\begin{aligned} \sin(2\arctan x)&=2\cdot\frac{x}{\sqrt{1+x^2}}\cdot\frac{1}{\sqrt{1+x^2}}\\ &=\boxed{\dfrac{2x}{1+x^{2}}}. \end{aligned}

The same triangle gives cos⁡(2arctan⁡x)=cos⁡2θ−sin⁡2θ=1−x21+x2\cos(2\arctan x)=\cos^2\theta-\sin^2\theta=\dfrac{1-x^2}{1+x^2}.

Tip

Pattern to remember. These are the tangent half-angle substitutions: with t=arctan⁡xt=\arctan x, sin⁡2t=2x1+x2\sin 2t=\dfrac{2x}{1+x^2} and cos⁡2t=1−x21+x2\cos 2t=\dfrac{1-x^2}{1+x^2}. Valid for all real xx because arctan⁡\arctan has domain (−∞,∞)(-\infty,\infty) and always lands in QI/QIV where the triangle sides carry the right signs.

The arctangent sum formula and the ab>1ab>1 trap

Concept
Adding two arctangents

The tangent addition formula gives

tan⁡(arctan⁡a+arctan⁡b)=a+b1−ab,\tan(\arctan a+\arctan b)=\frac{a+b}{1-ab},

so it is tempting to write arctan⁡a+arctan⁡b=arctan⁡ ⁣a+b1−ab\arctan a+\arctan b=\arctan\!\dfrac{a+b}{1-ab}. But arctan⁡\arctan only returns values in (−π2,π2)\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right), while the true sum can land outside it. The correct statement carries a quadrant correction:

arctan⁡a+arctan⁡b=arctan⁡ ⁣a+b1−ab+kπ,\arctan a+\arctan b=\arctan\!\frac{a+b}{1-ab}+k\pi,

where k=0k=0 if ab<1ab<1,  k=+1\ k=+1 if ab>1ab>1 and a>0a>0, and  k=−1\ k=-1 if ab>1ab>1 and a<0a<0.

Example
Safe case (ab<1ab<1) vs. the trap (ab>1ab>1)

(a) Safe. Evaluate arctan⁡12+arctan⁡13\arctan\tfrac12+\arctan\tfrac13. Here ab=16<1ab=\tfrac16<1, so no correction:

arctan⁡12+arctan⁡13=arctan⁡12+131−16=arctan⁡5/65/6=arctan⁡1=π4.\arctan\tfrac12+\arctan\tfrac13=\arctan\frac{\tfrac12+\tfrac13}{1-\tfrac16}=\arctan\frac{5/6}{5/6}=\arctan 1=\boxed{\tfrac{\pi}{4}}.

(b) Trap. Evaluate arctan⁡2+arctan⁡3\arctan 2+\arctan 3. Now ab=6>1ab=6>1, so the raw formula lies. Compute the fraction first:

2+31−2⋅3=5−5=−1,arctan⁡(−1)=−π4.\frac{2+3}{1-2\cdot3}=\frac{5}{-5}=-1,\qquad \arctan(-1)=-\tfrac{\pi}{4}.

That negative answer is impossible: arctan⁡2\arctan 2 and arctan⁡3\arctan 3 are both bigger than π4\tfrac{\pi}{4}, so their sum exceeds π2\tfrac{\pi}{2}. Since ab>1ab>1 and a>0a>0 we add π\pi:

arctan⁡2+arctan⁡3=−π4+π=3π4.\arctan 2+\arctan 3=-\tfrac{\pi}{4}+\pi=\boxed{\tfrac{3\pi}{4}}.

Check: 3π4\tfrac{3\pi}{4} is in QII where tangent is −1-1. ✓

Tip

Sanity test before trusting the formula. Estimate the size of each arctangent. If both inputs are positive and their sum of angles clearly exceeds π2\tfrac{\pi}{2}, you are in the ab>1ab>1 regime and must add π\pi. When in doubt, add angles numerically on a calculator to pick the right branch.

Telescoping (Machin-like) sums

Concept
Turning a sum into a difference

The difference version of the formula is often exact with k=0k=0:

arctan⁡1n−arctan⁡1n+1=arctan⁡11+n(n+1)=arctan⁡1n2+n+1.\arctan\frac{1}{n}-\arctan\frac{1}{n+1}=\arctan\frac{1}{1+n(n+1)}=\arctan\frac{1}{n^2+n+1}.

(The product of the two small inputs is 1n(n+1)<1\tfrac{1}{n(n+1)}<1, so no correction is needed.) A long sum of the right-hand terms then telescopes: consecutive pieces cancel and only the endpoints survive.

Example
Evaluate a telescoping arctangent sum

Find S=arctan⁡13+arctan⁡17+arctan⁡113+arctan⁡121\displaystyle S=\arctan\tfrac13+\arctan\tfrac17+\arctan\tfrac{1}{13}+\arctan\tfrac{1}{21}. Notice 3=12+1+13=1^2+1+1,  7=22+2+1\ 7=2^2+2+1,  13=32+3+1\ 13=3^2+3+1,  21=42+4+1\ 21=4^2+4+1. Rewrite each term as a difference:

S=(arctan⁡11−arctan⁡12)+(arctan⁡12−arctan⁡13)+(arctan⁡13−arctan⁡14)+(arctan⁡14−arctan⁡15).\begin{aligned} S&=\Bigl(\arctan\tfrac11-\arctan\tfrac12\Bigr)+\Bigl(\arctan\tfrac12-\arctan\tfrac13\Bigr)\\ &\quad+\Bigl(\arctan\tfrac13-\arctan\tfrac14\Bigr)+\Bigl(\arctan\tfrac14-\arctan\tfrac15\Bigr). \end{aligned}

Every interior term cancels, leaving only the ends:

S=arctan⁡1−arctan⁡15=π4−arctan⁡15≈0.588.S=\arctan 1-\arctan\tfrac15=\tfrac{\pi}{4}-\arctan\tfrac15\approx 0.588.

The same idea underlies Machin's formula π4=4arctan⁡15−arctan⁡1239\tfrac{\pi}{4}=4\arctan\tfrac15-\arctan\tfrac{1}{239}, historically used to compute π\pi.

Nested compositions with range traps

Concept
arccos⁡(cos⁡θ)\arccos(\cos\theta) is a “fold,” not the identity

Because arccos⁡\arccos must output a value in [0,π][0,\pi], the composition arccos⁡(cos⁡θ)\arccos(\cos\theta) returns the unique number in [0,π][0,\pi] having the same cosine as θ\theta:

arccos⁡(cos⁡θ)={θ,0≤θ≤π,2π−θ,π≤θ≤2π,−θ,−π≤θ≤0.\arccos(\cos\theta)= \begin{cases} \theta, & 0\le\theta\le\pi,\\[2pt] 2\pi-\theta, & \pi\le\theta\le 2\pi,\\[2pt] -\theta, & -\pi\le\theta\le 0. \end{cases}

Geometrically it folds the angle back into [0,π][0,\pi] using the even symmetry cos⁡θ=cos⁡(−θ)\cos\theta=\cos(-\theta) and periodicity.

Example
Same input, three wrappers

Take θ=4π3\theta=\dfrac{4\pi}{3} (which is not in [0,π][0,\pi]).

arccos⁡ ⁣(cos⁡4π3)=2π−4π3=2π3(fold into [0,π]),arcsin⁡ ⁣(sin⁡4π3)=arcsin⁡ ⁣(−32)=−π3(range [−π2,π2]),arctan⁡ ⁣(tan⁡4π3)=arctan⁡ ⁣(3)=π3(range (−π2,π2)).\begin{aligned} \arccos\!\left(\cos\tfrac{4\pi}{3}\right)&=2\pi-\tfrac{4\pi}{3}=\boxed{\tfrac{2\pi}{3}} &&\text{(fold into }[0,\pi]\text{)},\\ \arcsin\!\left(\sin\tfrac{4\pi}{3}\right)&=\arcsin\!\left(-\tfrac{\sqrt3}{2}\right)=\boxed{-\tfrac{\pi}{3}} &&\text{(range }\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]),\\ \arctan\!\left(\tan\tfrac{4\pi}{3}\right)&=\arctan\!\left(\sqrt3\right)=\boxed{\tfrac{\pi}{3}} &&\text{(range }\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)). \end{aligned}

The three “inverses of the same angle” disagree because each function reports the representative living in its own range. Always reduce the inside first, then land in the outer range.

Solving equations that contain inverse functions

Example
An equation in arctan⁡\arctan

Solve arctan⁡(x)+arctan⁡(2x)=π4\arctan(x)+\arctan(2x)=\dfrac{\pi}{4}. Take the tangent of both sides and use the sum formula on the left:

tan⁡(arctan⁡x+arctan⁡2x)=x+2x1−(x)(2x)=3x1−2x2,tan⁡π4=1.\tan\bigl(\arctan x+\arctan 2x\bigr)=\frac{x+2x}{1-(x)(2x)}=\frac{3x}{1-2x^2},\qquad \tan\tfrac{\pi}{4}=1.

So 3x1−2x2=1⇒3x=1−2x2⇒2x2+3x−1=0\dfrac{3x}{1-2x^2}=1\Rightarrow 3x=1-2x^2\Rightarrow 2x^2+3x-1=0, giving

x=−3±9+84=−3±174.x=\frac{-3\pm\sqrt{9+8}}{4}=\frac{-3\pm\sqrt{17}}{4}.

Check for extraneous roots. The two arctangents must add to a positive π4\tfrac{\pi}{4}, so we need x>0x>0. Only x=−3+174≈0.28x=\dfrac{-3+\sqrt{17}}{4}\approx0.28 works; the negative root ≈−1.78\approx-1.78 makes the left side negative (and is the spurious solution introduced by taking tan⁡\tan). Answer: x=−3+174\boxed{x=\dfrac{-3+\sqrt{17}}{4}}.

Tip

Taking tan⁡\tan/sin⁡\sin/cos⁡\cos can add fake solutions. Those operations are not one-to-one, so every algebraic root must be tested back in the original equation --- confirm both the value and the correct sign/quadrant of each inverse term.

Domain of a composite inverse expression

Concept
Chaining domain requirements

For a composition to be defined, the inner output must satisfy the outer function's domain and the inner function's own domain. For arcsin⁡\arcsin and arccos⁡\arccos the input must lie in [−1,1][-1,1]; arctan⁡\arctan accepts anything.

Example
Find the domain of f(x)=arccos⁡(2x−1)f(x)=\arccos(2x-1)

The outer arccos⁡\arccos demands its argument stay in [−1,1][-1,1]:

−1≤2x−1≤1  ⟹  0≤2x≤2  ⟹  0≤x≤1.-1\le 2x-1\le 1 \;\Longrightarrow\; 0\le 2x\le 2 \;\Longrightarrow\; \boxed{0\le x\le 1}.

So the domain is [0,1][0,1]. As a bonus, the range is still all of [0,π][0,\pi] because 2x−12x-1 sweeps the full interval [−1,1][-1,1] as xx runs over [0,1][0,1].

Tip

Two-sided inputs need two inequalities. A restriction like ∣u∣≤1|u|\le 1 is shorthand for −1≤u≤1-1\le u\le 1; solve both halves. For a nested radical such as arcsin⁡ ⁣(x)\arcsin\!\bigl(\sqrt{x}\bigr) you must also enforce the inner domain (x≥0x\ge0) and the outer one (x≤1\sqrt{x}\le1), giving 0≤x≤10\le x\le1.

Formulas, Proofs & Tips

Tip
Inverse trig ranges
arcsin⁡:[−π2,π2],arccos⁡:[0,π],arctan⁡:(−π2,π2)\arcsin: \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\qquad \arccos: [0,\pi],\qquad \arctan: \left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)

What it means. Each inverse returns one chosen angle, not every possible one.

Example. arcsin⁡(1)=π2\arcsin(1)=\tfrac\pi2 and arccos⁡(0)=π2\arccos(0)=\tfrac\pi2, both inside their allowed ranges.

Why it works. sin⁡\sin, cos⁡\cos and tan⁡\tan repeat, so they are not one-to-one and have no inverse until you restrict them. These intervals are the standard restrictions on which each function rises or falls exactly once, hitting every output.

Tip. arcsin⁡\arcsin never returns an angle in the second quadrant. If a problem's angle lives there, find the reference angle and adjust yourself.