Mathematics · Glossary

What is arcsin⁡, arccos⁡, arctan⁡?

Definition 4.9 University Mathematics — Year 1 · Chapter 4 — Standard Functions

The restrictions

sin ⁣:[π2,π2][1,1],cos ⁣:[0,π][1,1],tan ⁣:(π2,π2)R\sin \colon \intcc{-\tfrac\pi2}{\tfrac\pi2} \to \intcc{-1}{1}, \qquad \cos \colon \intcc{0}{\pi} \to \intcc{-1}{1}, \qquad \tan \colon \intoo{-\tfrac\pi2}{\tfrac\pi2} \to \R

are strictly monotonic bijections. Their inverse maps are written arcsin\arcsin, arccos\arccos and arctan\arctan. Thus, for instance, y=arcsinxy = \arcsin x (x[1,1]x \in \intcc{-1}{1}) is the angle in [π2,π2]\intcc{-\frac\pi2}{\frac\pi2} whose sine is xx.

From left to right: , , . Each inherits its graph from the restricted direct function by reflection in the line y = x.
From left to right: arcsin\arcsin, arccos\arccos, arctan\arctan. Each inherits its graph from the restricted direct function by reflection in the line y=xy = x.

Examples

Example 4.11 (Recognizing a hidden constant)

Study g(x)=arctan1x1+xg(x) = \arctan\dfrac{1 - x}{1 + x} on (1,+)\intoo{-1} {+\infty}. Chain rule and a short computation:

g(x)=11+(1x1+x)2(1+x)(1x)(1+x)2=2(1+x)2+(1x)2=11+x2,g'(x) = \frac{1}{1 + \bigl(\frac{1-x}{1+x}\bigr)^2}\cdot \frac{-(1+x) - (1-x)}{(1+x)^2} = \frac{-2}{(1+x)^2 + (1-x)^2} = \frac{-1}{1 + x^2} ,

since (1+x)2+(1x)2=2+2x2(1+x)^2 + (1-x)^2 = 2 + 2x^2. So g=arctang' = -\arctan': the function g+arctang + \arctan has zero derivative on the interval (1,+)\intoo{-1}{+\infty}, hence is constant there; its value at x=0x = 0 is arctan1+arctan0=π4\arctan 1 + \arctan 0 = \frac\pi4. Conclusion:

arctan1x1+x=π4arctanx(x>1).\arctan\frac{1 - x}{1 + x} = \frac\pi4 - \arctan x \qquad (x > -1) .

On (,1)\intoo{-\infty}{-1} the same derivative computation holds but the constant is different (3π4-\frac{3\pi}4: evaluate the limit as xx \to -\infty). The insight: “zero derivative implies constant” is an interval-by-interval statement — exactly the subtlety exploited in Exercise 4.6 and Exercise 4.10.

Example 4.14 (Wrapping back to the principal interval)

Compute

arctan(tan3π4)andarctan(tan17π5).\arctan\Bigl(\tan\frac{3\pi}4\Bigr) \qquad\text{and}\qquad \arctan\Bigl(\tan\frac{17\pi}5\Bigr).

The recipe: replace the angle by the unique angle of (π2,π2)\intoo{-\frac\pi2}{\frac\pi2} with the same tangent, i.e. subtract the right multiple of π\pi (the period of tan\tan). First: 3π4π=π4\frac{3\pi}4 - \pi = -\frac\pi4, so the answer is π4-\frac\pi4. Second: 17π53π=2π5(π2,π2)\frac{17\pi} 5 - 3\pi = \frac{2\pi}5 \in \intoo{-\frac\pi2}{\frac\pi2}, so the answer is 2π5\frac{2\pi}5. The computation is a Euclidean division of the angle by π\pi in disguise — and the analogous recipes for arcsin\arcsin (reflect into [π2,π2]\intcc{-\frac\pi2}{\frac\pi2}, period 2π2\pi) and arccos\arccos (reflect into [0,π]\intcc0\pi) drive Exercise 4.1 and the piecewise answer of Exercise 4.10.

Example 4.15 (Adding arctangents safely)

Let us prove

arctan12+arctan15+arctan18=π4.\arctan\frac12 + \arctan\frac15 + \arctan\frac18 = \frac\pi4 .

Two ingredients: the tangent addition formula, and — the step beginners forget — a localization of the sum. First, tan(u+v)=tanu+tanv1tanutanv\tan(u + v) = \frac{\tan u + \tan v}{1 - \tan u\tan v} with u=arctan12u = \arctan\frac12, v=arctan15v = \arctan\frac15 gives

tan(u+v)=12+151110=7/109/10=79,thentan(u+v+arctan18)=79+181772=65/7265/72=1.\tan(u + v) = \frac{\frac12 + \frac15}{1 - \frac1{10}} = \frac{7/10}{9/10} = \frac79, \qquad\text{then}\qquad \tan\Bigl(u + v + \arctan\frac18\Bigr) = \frac{\frac79 + \frac18}{1 - \frac7{72}} = \frac{65/72}{65/72} = 1 .

Second, each of the three angles lies in (0,π4)\intoo0{\frac\pi4} (their arguments are less than 11), so the sum lies in (0,3π4)\intoo0{\frac{3\pi}4}; the only angle there with tangent 11 is π4\frac\pi4. Without the localization the conclusion would be “π4\frac\pi4 up to a multiple of π\pi” — half a proof. The same two-step discipline runs Exercise 4.8 and Exercise 4.12.

Read in context →