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.

Because the trig functions repeat, they have inverses only on restricted domains: arcsin⁡\arcsin returns angles in [−π2,π2][-\tfrac\pi2, \tfrac\pi2], arccos⁡\arccos in [0,π][0, \pi], arctan⁡\arctan in (−π2,π2)(-\tfrac\pi2, \tfrac\pi2). Compositions like cos⁡(arcsin⁡x)\cos(\arcsin x) are best evaluated by drawing the right triangle the inner function describes.

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).

domainrangearcsin⁡x[−1,1][−π2,π2]arccos⁡x[−1,1][0,π]arctan⁡xall reals(−π2,π2)\begin{array}{c|c|c} & \text{domain} & \text{range}\\ \hline \arcsin x & [-1,1] & \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\\ \arccos x & [-1,1] & [0,\pi]\\ \arctan x & \text{all reals} & \left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right) \end{array}

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).

A function can be undone only if no output comes from two different inputs, and sine fails badly: sin⁡x=12\sin x=\tfrac12 at π6\tfrac{\pi}{6}, 5π6\tfrac{5\pi}{6}, and infinitely many more. So we keep one stretch of the graph that passes the horizontal-line test and still reaches every value from −1-1 to 11. For sine that stretch is [−π2,π2]\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right], where it rises steadily; cosine falls steadily on [0,π][0,\pi]. Whatever stretch is kept becomes the set of answers the inverse is allowed to give.

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.

Side note
Where it goes wrong

Writing arcsin⁡(sin⁡3π4)=3π4\arcsin(\sin\tfrac{3\pi}{4}) = \tfrac{3\pi}{4}. Arcsine only returns angles between −π2-\tfrac\pi2 and π2\tfrac\pi2; the answer is π4\tfrac\pi4.

Try it
Quick check

Which of these are defined? Give the exact value where there is one: arcsin⁡2\arcsin 2,  arccos⁡(−1)\ \arccos(-1),  arctan⁡100\ \arctan 100.

Show the answer

Answer: arcsin⁡2\arcsin 2 is undefined; arccos⁡(−1)=π\arccos(-1)=\pi; arctan⁡100\arctan 100 is defined, about 1.56081.5608. No angle has sine 22, so 22 is outside the domain [−1,1][-1,1]. The angle in [0,π][0,\pi] with cosine −1-1 is π\pi. Arctangent accepts every real number; a large input gives an angle just under π2≈1.5708\tfrac{\pi}{2}\approx 1.5708.

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
range of arccos3π/4−√2/2
Arccosine answers from the upper half-circle, [0,π][0,\pi]: the point with x=−22x=-\tfrac{\sqrt2}{2} there is at 3π4\tfrac{3\pi}{4}.

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.

Try it
Quick check

Evaluate exactly: arctan⁡(−3)\arctan(-\sqrt3) and arccos⁡32\arccos\dfrac{\sqrt3}{2}.

Show the answer

Answer: −π3-\dfrac{\pi}{3} and π6\dfrac{\pi}{6}. tan⁡π3=3\tan\dfrac{\pi}{3}=\sqrt3, and arctangent answers in (−π2,π2)\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right), so a negative input gives the negative angle −π3-\tfrac{\pi}{3} (not 2π3\tfrac{2\pi}{3}, which also has tangent −3-\sqrt3 but is out of range). cos⁡π6=32\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2} and π6∈[0,π]\tfrac{\pi}{6}\in[0,\pi].

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.

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

The triangle is a legal shortcut because the inner inverse returns an angle whose sine (or cosine, or tangent) is known, and any right triangle with that ratio has that angle. Only the sign needs care: arcsine and arctangent of a negative number land in Quadrant IV, where cosine is still positive, and arccosine of a negative number lands in Quadrant II, where sine is still positive.

Example
Numeric composition
θ435
sin⁡θ=35\sin\theta=\tfrac35 fixes opposite 33 and hypotenuse 55; the adjacent leg is 44.

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. Draw that triangle (above). 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
θx√(1 − x²)1
For θ=arccos⁡x\theta=\arccos x: adjacent xx, hypotenuse 11.

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}}}.

Side note
Where it goes wrong

Copying the sign of the inside. For cos⁡ ⁣(arcsin⁡ ⁣(−35))\cos\!\left(\arcsin\!\left(-\tfrac35\right)\right) it is tempting to answer −45-\tfrac45. But arcsin⁡ ⁣(−35)\arcsin\!\left(-\tfrac35\right) is a Quadrant IV angle, between −π2-\tfrac{\pi}{2} and 00, and cosine is positive there, so the answer is +45+\tfrac45. The triangle gives the size; the range of the inner inverse gives the sign.

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.

Try it
Quick check

(a) Evaluate sin⁡ ⁣(arctan⁡512)\sin\!\left(\arctan\dfrac{5}{12}\right). (b) Write cos⁡(arctan⁡x)\cos(\arctan x) as an algebraic expression.

Show the answer

(a) 513\dfrac{5}{13}. Opposite 55, adjacent 1212, so the hypotenuse is 1313 and sin⁡θ=513\sin\theta=\tfrac{5}{13}. (b) 11+x2\dfrac{1}{\sqrt{1+x^2}}. Opposite xx, adjacent 11, hypotenuse 1+x2\sqrt{1+x^2}, so cos⁡θ=11+x2\cos\theta=\dfrac{1}{\sqrt{1+x^2}}. It is positive for every xx, as it must be: arctan⁡x\arctan x always lies in Quadrant I or IV.

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.

Example
The classic trap
5π/6π/6range of arcsin
Same height, same sine; only π6\tfrac{\pi}{6} is on the right half-circle where arcsine answers.

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
7π/65π/6
Same xx-coordinate, same cosine; 5π6\tfrac{5\pi}{6} is the one in [0,π][0,\pi].

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].

Try it
Quick check

Evaluate arccos⁡ ⁣(cos⁡5π3)\arccos\!\left(\cos\dfrac{5\pi}{3}\right) and arctan⁡ ⁣(tan⁡3π4)\arctan\!\left(\tan\dfrac{3\pi}{4}\right).

Show the answer

Answer: π3\dfrac{\pi}{3} and −π4-\dfrac{\pi}{4}. cos⁡5π3=12\cos\dfrac{5\pi}{3}=\tfrac12, and the angle in [0,π][0,\pi] with cosine 12\tfrac12 is π3\tfrac{\pi}{3}. tan⁡3π4=−1\tan\dfrac{3\pi}{4}=-1, and the angle in (−π2,π2)\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right) with tangent −1-1 is −π4-\tfrac{\pi}{4}. Neither outer function simply cancels, because neither starting angle is in its range.

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.

Swapping inputs and outputs swaps the xx- and yy-coordinates of every point, and the map (x,y)↦(y,x)(x,y)\mapsto(y,x) is the reflection in y=xy=x. The restricted sine runs from (−π2,−1)\left(-\tfrac{\pi}{2},-1\right) to (π2,1)\left(\tfrac{\pi}{2},1\right), so arcsine runs from (−1,−π2)\left(-1,-\tfrac{\pi}{2}\right) to (1,π2)\left(1,\tfrac{\pi}{2}\right). Tangent's vertical asymptotes x=±π2x=\pm\tfrac{\pi}{2} reflect into arctangent's horizontal asymptotes y=±π2y=\pm\tfrac{\pi}{2}.

Tip
xy−6−4−224π/2−π/2
y=arctan⁡xy=\arctan x creeps toward y=±π2y=\pm\tfrac{\pi}{2} but never reaches them.

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.

Try it
Quick check

Give the domain and range of y=2arcsin⁡xy=2\arcsin x and of y=arctan⁡x+π2y=\arctan x+\dfrac{\pi}{2}.

Show the answer

Answer: y=2arcsin⁡xy=2\arcsin x has domain [−1,1][-1,1] and range [−π,π][-\pi,\pi]; y=arctan⁡x+π2y=\arctan x+\tfrac{\pi}{2} has domain all reals and range (0,π)(0,\pi). The factor 22 stretches outputs only, doubling [−π2,π2]\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]. Adding π2\tfrac{\pi}{2} lifts the open interval (−π2,π2)\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right) to (0,π)(0,\pi); the endpoints stay excluded because the asymptotes are never reached.

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
2π/34π/3x = −1/2
The vertical line x=−12x=-\tfrac12 cuts the circle twice: two solutions per turn.

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).

Try it
Quick check

Solve sin⁡x=−22\sin x=-\dfrac{\sqrt2}{2} on [0,2π)[0,2\pi).

Show the answer

Answer: x=5π4x=\dfrac{5\pi}{4} and x=7π4x=\dfrac{7\pi}{4}. The reference angle is arcsin⁡22=π4\arcsin\dfrac{\sqrt2}{2}=\dfrac{\pi}{4}. Sine is negative in Quadrants III and IV, giving π+π4\pi+\dfrac{\pi}{4} and 2π−π42\pi-\dfrac{\pi}{4}. (The calculator's arcsin⁡ ⁣(−22)=−π4\arcsin\!\left(-\tfrac{\sqrt2}{2}\right)=-\tfrac{\pi}{4} is coterminal with 7π4\tfrac{7\pi}{4}, and misses 5π4\tfrac{5\pi}{4} entirely.)

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

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)
θ1x√(1 + x²)
For θ=arctan⁡x\theta=\arctan x: opposite xx, adjacent 11.

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} (above). 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.

Why It Works

A function has an inverse only if it is one-to-one, and a periodic function never is — so we choose one piece where it is: the piece of sine from −π2-\tfrac\pi2 to π2\tfrac\pi2 is increasing and covers every value in [−1,1][-1, 1] exactly once. On that piece sin⁡(arcsin⁡x)=x\sin(\arcsin x) = x always, but arcsin⁡(sin⁡θ)=θ\arcsin(\sin\theta) = \theta only for θ\theta in the chosen range. Drawing the triangle for arcsin⁡x\arcsin x (opposite xx, hypotenuse 11, adjacent 1−x2\sqrt{1 - x^2}) makes cos⁡(arcsin⁡x)=1−x2\cos(\arcsin x) = \sqrt{1 - x^2} visible.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

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.