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Topos Uranos
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The inverse trigonometric functions

Summary

This lesson constructs the inverses of the sine, the cosine and the tangent. Since these functions are periodic, they are not injective, and to invert them one must restrict them to an interval on which they take each value only once: the sine to [−π2,π2]\icc{-\frac{\pi}{2}}{\frac{\pi}{2}}, the cosine to [0,π]\icc{0}{\pi} and the tangent to (−π2,π2)\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}. From the axiom of traversal by arc length, which states that the abscissa of the traversed point strictly decreases along the first half turn, it deduces that the cosine is strictly decreasing on [0,π]\icc{0}{\pi}; from this and from the symmetries it proves that the sine and the tangent are strictly increasing on their intervals, and that the three restrictions are bijective; it defines the arcsine, the arccosine and the arctangent, and obtains their graphs as reflections in the line y=xy = x. It then proves the identities that relate them, carefully distinguishes sin⁡(arcsin⁡x)=x\sin(\arcsin x) = x, which always holds, from arcsin⁡(sin⁡x)=x\arcsin(\sin x) = x, which holds only on the principal interval, and solves the equations sin⁡x=c\sin x = c, cos⁡x=c\cos x = c and tan⁡x=c\tan x = c over one full turn.

Learning objectives

  1. Deduce from the traversal axiom the strict monotonicity of the cosine and of the sine on their principal intervals, use it to prove that of the tangent, and define the three inverses with their domains and ranges.
  2. Construct the graphs of the arcsine, the arccosine and the arctangent as reflections in the line y=xy = x, and compute their values at the special points.
  3. Prove the composition, symmetry and complementary angle identities of the inverse functions, and simplify expressions that combine trigonometric functions with their inverses.
  4. Solve the elementary trigonometric equations on the interval [0,2π)\ico{0}{2\pi} by means of the inverse functions and the symmetries of the unit circle.

The restriction of the trigonometric functions

According to the Course I lesson on function operations, a function has an inverse if and only if it is bijective, and then the graph of the inverse is the reflection of its own graph in the line y=xy = x. Now, the trigonometric functions studied in the lesson on the trigonometric functions and their graphs are not injective: the sine and the cosine have period 2π2\pi, and the tangent has period π\pi, so that each value is repeated infinitely many times. Moreover, within a single period the values repeat: by the symmetry of supplementary angles, proved in the lesson on trigonometric ratios, sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin x holds. If this is so, to invert these functions one must restrict them to an interval on which they are injective and on which they take all their values. The intervals chosen contain the acute angles, and on them the function is strictly monotonic.

TheoremMonotonicity on the principal intervals

The cosine is strictly decreasing on [0,π]\icc{0}{\pi}; the sine is strictly increasing on [−π2,π2]\icc{-\frac{\pi}{2}}{\frac{\pi}{2}}, and the tangent is strictly increasing on (−π2,π2)\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}.

We use the definition of the sine and the cosine by means of the point P(θ)=(cos⁡θ,sin⁡θ)P(\theta) = \left( \cos \theta, \sin \theta \right) of the unit circle, from the lesson on trigonometric ratios, and the axiom of traversal by arc length, admitted in the lesson on angles and their measure. The monotonicity of the cosine is a restatement of the half-turn property of that axiom; that of the sine and that of the tangent are proved from it.

Proof

  1. 0≤α<β≤π0 \leq \alpha < \beta \leq \pi

    Let α\alpha and β\beta satisfy 0≤α<β≤π0 \leq \alpha < \beta \leq \pi. The points P(α)P(\alpha) and P(β)P(\beta) are the endpoints of the arcs of lengths α\alpha and β\beta that start from (1,0)(1, 0) counterclockwise, and both lie on the first half turn of the traversal, which covers the upper semicircle.

  2. 0≤α<β≤π→cos⁡α>cos⁡β\dato{c}{0 \leq \alpha < \beta \leq \pi \rightarrow \cos \alpha > \cos \beta}

    By the half-turn property of the axiom, during the first half turn the abscissa of the traversed point strictly decreases: that of P(α)P(\alpha) is greater than that of P(β)P(\beta). Since the cosine is, by definition, the abscissa of P(θ)P(\theta), the inequality between the abscissas is an inequality between cosines.

  3. sin⁡α=cos⁡(π2−α)<cos⁡(π2−β)=sin⁡β\sin \alpha = \resaltar{\cos\left( \frac{\pi}{2} - \alpha \right)} < \resaltar{\cos\left( \frac{\pi}{2} - \beta \right)} = \dato{s}{\sin \beta}

    Now let −π2≤α<β≤π2-\frac{\pi}{2} \leq \alpha < \beta \leq \frac{\pi}{2}. Their complements satisfy π2−β<π2−α\frac{\pi}{2} - \beta < \frac{\pi}{2} - \alpha and lie in [0,π]\icc{0}{\pi}; by the symmetry of complementary angles, the sine of an angle is the cosine of its complement, and since the cosine decreases on [0,π]\icc{0}{\pi}, the sine increases on [−π2,π2]\icc{-\frac{\pi}{2}}{\frac{\pi}{2}}.

  4. tan⁡α=sin⁡αcos⁡α<sin⁡βcos⁡α<sin⁡βcos⁡β=tan⁡β\tan \alpha = \frac{\sin \alpha}{\cos \alpha} < \frac{\resaltar{\sin \beta}}{\cos \alpha} < \frac{\sin \beta}{\resaltar{\cos \beta}} = \dato{t}{\tan \beta}

    For the tangent, let 0≤α<β<π20 \leq \alpha < \beta < \frac{\pi}{2}. By the preceding steps, 0≤sin⁡α<sin⁡β0 \leq \sin \alpha < \sin \beta and cos⁡α>cos⁡β>cos⁡π2=0\cos \alpha > \cos \beta > \cos \frac{\pi}{2} = 0. Increasing first the numerator and then decreasing the denominator, both positive, increases the quotient.

  5. tan⁡α=−tan⁡(−α)<−tan⁡(−β)=tan⁡β\tan \alpha = \resaltar{-\tan(-\alpha)} < \resaltar{-\tan(-\beta)} = \tan \beta

    The tangent is odd, because it is the quotient of the sine, which is odd, by the cosine, which is even. If −π2<α<β≤0-\frac{\pi}{2} < \alpha < \beta \leq 0, the opposites satisfy 0≤−β<−α0 \leq -\beta < -\alpha, and changing sign reverses the inequality of the preceding step; if α<0<β\alpha < 0 < \beta, the tangent is negative at α\alpha and positive at β\beta. Therefore the tangent increases on the whole interval (−π2,π2)\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}.

The bijective restrictions

Every strictly monotonic function is injective, by the theorem on monotonic functions in the lesson on function operations. It remains to show that each restriction takes all the values of its range.

TheoremBijective restrictions

The sine, restricted to [−π2,π2]\icc{-\frac{\pi}{2}}{\frac{\pi}{2}}, is a bijection onto [−1,1]\icc{-1}{1}; the cosine, restricted to [0,π]\icc{0}{\pi}, is a bijection onto [−1,1]\icc{-1}{1}, and the tangent, restricted to (−π2,π2)\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}, is a bijection onto R\mathbb{R}.

Injectivity follows from the preceding theorem. That each restriction takes all its values follows from the covering property of the traversal axiom, according to which every point of the unit circle is the endpoint of an arc that starts from (1,0)(1, 0), and from the lemma according to which one turn does not repeat points, both from the lesson on angles and their measure.

Proof

  1. 0≤α<β≤π→cos⁡α≠cos⁡β0 \leq \alpha < \beta \leq \pi \rightarrow \resaltar{\cos \alpha \neq \cos \beta}

    By the preceding theorem, the three restrictions are strictly monotonic, and every strictly monotonic function is injective: two distinct angles of the upper semicircle have distinct abscissas.

  2. Q=(c,1−c2),c2+(1−c2)2=1→Q=P(θ)Q = \left( c, \sqrt{1 - c^2} \right), \quad \resaltar{c^2 + \left( \sqrt{1 - c^2} \right)^2 = 1} \rightarrow \dato{q}{Q = P(\theta)}

    Let cc satisfy −1≤c≤1-1 \leq c \leq 1. The point QQ below lies on the unit circle; by the covering property and because one turn does not repeat points, it is P(θ)P(\theta) for a unique θ\theta in [0,2π)\ico{0}{2\pi}.

  3. sin⁡θ=1−c2≥0→θ∈[0,π],cos⁡θ=c\sin \theta = \sqrt{1 - c^2} \geq 0 \rightarrow \theta \in \icc{0}{\pi}, \quad \dato{c}{\cos \theta = c}

    The ordinate of QQ is not negative, and the sine is negative on (π,2π)\ioo{\pi}{2\pi}, according to the signs by quadrant from the lesson on trigonometric ratios: that θ\theta lies in [0,π]\icc{0}{\pi}, and its cosine is cc. The restricted cosine takes all the values of [−1,1]\icc{-1}{1}.

  4. sin⁡(π2−θ)=cos⁡θ=c,π2−θ∈[−π2,π2]\sin\left( \frac{\pi}{2} - \theta \right) = \resaltar{\cos \theta} = \resaltar{c}, \quad \frac{\pi}{2} - \theta \in \icc{-\frac{\pi}{2}}{\frac{\pi}{2}}

    By the symmetry of complementary angles, the sine of π2−θ\frac{\pi}{2} - \theta is the cosine of θ\theta, and as θ\theta runs through [0,π]\icc{0}{\pi}, its complement runs through [−π2,π2]\icc{-\frac{\pi}{2}}{\frac{\pi}{2}}: the restricted sine also takes all the values of [−1,1]\icc{-1}{1}.

  5. tan⁡x=m,x−kπ∈(−π2,π2),tan⁡(x−kπ)=m\tan x = m, \quad x - k\pi \in \ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}, \quad \tan(x - k\pi) = m

    For the tangent, the theorem on its properties, from the lesson on the trigonometric functions and their graphs, says that its range is R\mathbb{R}: given mm, the line of slope mm through the origin meets the right semicircle at a point P(x)P(x) with xx between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}. It can also be seen by periodicity: if tan⁡x=m\tan x = m, subtracting from xx an integer multiple of π\pi, as in the reduction to the first turn, yields a number of that interval with the same tangent.