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Topos Uranos
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Sum and difference formulas, double angle and half angle

Summary

This lesson proves the formulas that express the sine, the cosine and the tangent of a sum or of a difference of angles in terms of the ratios of each angle. The first of them, that of the cosine of a difference, is obtained by computing in two ways the length of a chord of the unit circle: with the distance formula in Cartesian coordinates and with the distance formula in polar coordinates, which shows that the chord depends only on the difference of the angles. From it, using parity and the symmetries of complementary angles, the remaining sum and difference formulas are deduced, and from these, the double-angle formulas, the power-reducing formulas, the half-angle formulas, with the sign fixed by the quadrant, and the triple-angle formulas. With them, exact values are computed, such as those of π12\frac{\pi}{12} and π5\frac{\pi}{5}. Finally, the substitution t=tan⁡θ2t = \tan \frac{\theta}{2} expresses the sine and the cosine as rational fractions in tt, gives a rational parametrization of the unit circle and, with it, the formula that generates all Pythagorean triples.

Learning objectives

  1. Prove the formula for the cosine of a difference by means of the length of a chord of the unit circle, and deduce from it those for the sine and the tangent of a sum or difference, with their conditions of validity.
  2. Deduce the double-angle, power-reducing, half-angle and triple-angle formulas, and choose the sign of a square root according to the quadrant of the angle.
  3. Compute exact values of the trigonometric ratios of angles that are not special, such as π12\frac{\pi}{12}, 5π12\frac{5\pi}{12} and π5\frac{\pi}{5}, reducing them to polynomial equations when necessary.
  4. Express the sine and the cosine in terms of the tangent of the half angle, obtain the rational parametrization of the unit circle and use it to generate all Pythagorean triples.

The cosine of a difference

The symmetries proved in the lesson on trigonometric ratios and the unit circle, such as cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos \theta or sin⁡(π2−θ)=cos⁡θ\sin\left( \frac{\pi}{2} - \theta \right) = \cos \theta, relate the ratios of an angle to those of another angle obtained by adding a quadrantal angle to it or subtracting one from it. Now, these equalities are particular cases of a general problem: computing the ratios of α+β\alpha + \beta and of α−β\alpha - \beta from those of α\alpha and of β\beta, whatever these numbers may be. It should be noted from the outset that the ratios do not distribute over the sum: sin⁡(π6+π3)=sin⁡π2=1\sin\left( \frac{\pi}{6} + \frac{\pi}{3} \right) = \sin \frac{\pi}{2} = 1, whereas sin⁡π6+sin⁡π3=1+32\sin \frac{\pi}{6} + \sin \frac{\pi}{3} = \frac{1 + \sqrt{3}}{2}, which is greater than 11. The correct formula mixes sines and cosines, and the key to obtaining it is a chord of the unit circle.

As in the axiom of traversal by arc length from the lesson on angles and their measure, P(θ)P(\theta) denotes the point of the unit circle reached by traversing from (1,0)(1, 0) an arc of length ∣θ∣\abs{\theta}, counterclockwise if θ>0\theta > 0 and clockwise if θ<0\theta < 0; by the definition in the lesson cited, P(θ)=(cos⁡θ,sin⁡θ)P(\theta) = \left( \cos \theta, \sin \theta \right).

TheoremCosine of a difference

For all real numbers α\alpha and β\beta, cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta holds.

The distance between P(α)P(\alpha) and P(β)P(\beta) is computed first with their Cartesian coordinates and then with their polar coordinates, which are (1,α)(1, \alpha) and (1,β)(1, \beta); equating the two results yields the formula.

Proof

  1. d2=(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2d^2 = \left( \resaltar{\cos \alpha - \cos \beta} \right)^2 + \left( \resaltar{\sin \alpha - \sin \beta} \right)^2

    Let dd be the distance between P(α)P(\alpha) and P(β)P(\beta). By the distance formula from the Course I lesson on the equation of a line, its square is the sum of the squares of the differences of the coordinates.

  2. d2=(cos⁡2α+sin⁡2α)+(cos⁡2β+sin⁡2β)−2cos⁡αcos⁡β−2sin⁡αsin⁡β=2−2(cos⁡αcos⁡β+sin⁡αsin⁡β)d^2 = \left( \resaltar{\cos^2 \alpha + \sin^2 \alpha} \right) + \left( \resaltar{\cos^2 \beta + \sin^2 \beta} \right) - 2 \cos \alpha \cos \beta - 2 \sin \alpha \sin \beta = \dato{c}{2 - 2 \left( \cos \alpha \cos \beta + \sin \alpha \sin \beta \right)}

    We expand the two squares and group the sine and the cosine of each angle; by the Pythagorean identity, each group equals 11.

  3. d2=1+1−2⋅1⋅1⋅cos⁡(α−β)=2−2cos⁡(α−β)d^2 = 1 + 1 - 2 \cdot 1 \cdot 1 \cdot \resaltar{\cos(\alpha - \beta)} = \dato{p}{2 - 2 \cos(\alpha - \beta)}

    By the theorem on distance in polar coordinates, from the lesson on polar coordinates, applied with r1=r2=1r_1 = r_2 = 1, the same distance depends only on the difference of the angles. Geometrically, the rotation through the angle −β-\beta about the origin takes P(β)P(\beta) to (1,0)(1, 0) and P(α)P(\alpha) to P(α−β)P(\alpha - \beta), and it preserves distances: the two chords in the figure are equal.

  4. 2−2cos⁡(α−β)=2−2(cos⁡αcos⁡β+sin⁡αsin⁡β)↔cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\resaltar{2 - 2 \cos(\alpha - \beta)} = \resaltar{2 - 2 \left( \cos \alpha \cos \beta + \sin \alpha \sin \beta \right)} \leftrightarrow \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta

    We equate the two expressions for d2d^2, subtract 22 and divide by −2-2.

The proof requires no condition on α\alpha and β\beta: the theorem on distance in polar coordinates holds for arbitrary angles, and if P(α)=P(β)P(\alpha) = P(\beta) both expressions for d2d^2 are zero. Nor is there a vicious circle: that theorem was proved with the law of cosines, which in turn was obtained by placing the triangle in a coordinate system, without using angle sum formulas. The path can also be traveled in reverse: if the sides bb and cc of a triangle leave the origin with inclinations β\beta and α\alpha, the square of the distance between their endpoints is b2+c2−2bc(cos⁡αcos⁡β+sin⁡αsin⁡β)b^2 + c^2 - 2bc \left( \cos \alpha \cos \beta + \sin \alpha \sin \beta \right), and whoever has the formula for the cosine of a difference obtains from this the law of cosines from the lesson on solving triangles.