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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=arcsin⁡xy = \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)=arctan⁡1−x1+xg(x) = \arctan\dfrac{1 - x}{1 + x} on (−1, +∞)\intoo{-1} {+\infty}. Chain rule and a short computation:

g′(x)=11+(1−x1+x)2⋅−(1+x)−(1−x)(1+x)2=−2(1+x)2+(1−x)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+(1−x)2=2+2x2(1+x)^2 + (1-x)^2 = 2 + 2x^2. So g′=−arctan⁡′g' = -\arctan': the function g+arctan⁡g + \arctan has zero derivative on the interval (−1, +∞)\intoo{-1}{+\infty}, hence is constant there; its value at x=0x = 0 is arctan⁡1+arctan⁡0=π4\arctan 1 + \arctan 0 = \frac\pi4. Conclusion:

arctan⁡1−x1+x=π4−arctan⁡x(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 x→−∞x \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⁡(tan⁡3π4)andarctan⁡(tan⁡17π5).\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: 3π4−π=−π4\frac{3\pi}4 - \pi = -\frac\pi4, so the answer is −π4-\frac\pi4. Second: 17π5−3π=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

arctan⁡12+arctan⁡15+arctan⁡18=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)=tan⁡u+tan⁡v1−tan⁡utan⁡v\tan(u + v) = \frac{\tan u + \tan v}{1 - \tan u\tan v} with u=arctan⁡12u = \arctan\frac12, v=arctan⁡15v = \arctan\frac15 gives

tan⁡(u+v)=12+151−110=7/109/10=79,thentan⁡(u+v+arctan⁡18)=79+181−772=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.

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