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Topos Uranos
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Trigonometric ratios and the unit circle

Summary

This lesson defines the sine, the cosine and the tangent of an angle in two ways. The first, as quotients of the sides of a right triangle, applies to acute angles, and it is proved by similarity that it does not depend on the triangle chosen. The second, as the coordinates of the point where the terminal side of the angle meets the unit circle, applies to every real number and agrees with the first on acute angles. From it follow the signs in each quadrant, the Pythagorean identity, the special values of the angles of π6\frac{\pi}{6}, π4\frac{\pi}{4} and π3\frac{\pi}{3}, which are obtained from half an equilateral triangle and half a square, and the symmetries of the sine and the cosine, which are proved by reflecting the point on the circle and allow the ratios of any angle to be computed by means of its reference angle.

Learning objectives

  1. Define the six trigonometric ratios of an acute angle in a right triangle and prove, by similarity, that they do not depend on the triangle.
  2. Define the sine, the cosine and the other ratios of an arbitrary angle by means of the unit circle, recognize the tangent as the slope of the terminal side and prove that both definitions agree on acute angles.
  3. Prove the Pythagorean identity, the special values and the symmetries of the sine and the cosine, and use them to compute the ratios of any angle with the reference angle method.
  4. Obtain all the ratios of an angle from one of them and the quadrant, and apply the ratios to problems of heights, distances and areas.

The ratios of an acute angle

In a right triangle, the side opposite the right angle is called the hypotenuse, and the other two, the legs. The interior angles of a triangle add up to a straight angle, which measures π\pi; this fact of elementary geometry was admitted in the lesson on angles and their measure. If this is so, the two non-right angles of a right triangle add up to π2\frac{\pi}{2}, and each of them is acute, that is, it measures between 00 and π2\frac{\pi}{2}. With respect to one of these acute angles, of measure α\alpha, one of the legs is the opposite side, which does not touch its vertex, and the other is the adjacent side, which forms with the hypotenuse the sides of the angle.

Let α\alpha be an acute angle of a right triangle with hypotenuse cc, whose side opposite α\alpha measures aa and whose adjacent side measures bb. The sine, the cosine and the tangent of α\alpha are, respectively, the quotient of the opposite side by the hypotenuse, of the adjacent side by the hypotenuse, and of the opposite side by the adjacent one.
sin⁡α=ac,cos⁡α=bc,tan⁡α=ab\sin \alpha = \frac{a}{c}, \qquad \cos \alpha = \frac{b}{c}, \qquad \tan \alpha = \frac{a}{b}

Their reciprocals are called the cosecant, the secant and the cotangent: csc⁡α=ca\csc \alpha = \frac{c}{a}, sec⁡α=cb\sec \alpha = \frac{c}{b} and cot⁡α=ba\cot \alpha = \frac{b}{a}. The six are called trigonometric ratios, because they are ratios, that is, quotients, of the sides of a triangle. The definition, however, raises a doubt: there are infinitely many right triangles with an acute angle of measure α\alpha, large and small, and if each gave a different quotient, the words “the sine of α\alpha” would not designate any number. The following theorem dispels the doubt.

TheoremThe ratios do not depend on the triangle

If two right triangles have an acute angle of the same measure α\alpha, the quotient of any two sides of the first is equal to the quotient of the corresponding sides of the second. Consequently, the sine, the cosine and the tangent of α\alpha, and their reciprocals, depend only on α\alpha. Moreover, 0<sin⁡α<10 < \sin \alpha < 1 and 0<cos⁡α<10 < \cos \alpha < 1.

Consider two right triangles with an acute angle α\alpha, the first with legs a1a_1 and b1b_1 and hypotenuse c1c_1, and the second with legs a2a_2 and b2b_2 and hypotenuse c2c_2, where a1a_1 and a2a_2 are the legs opposite α\alpha. The similarity criteria for triangles of elementary geometry are admitted, the same ones that, in the Course I lesson on the equation of a line, guaranteed that the slope does not depend on the points chosen.

Proof

  1. a2=ka1,b2=kb1,c2=kc1\dato{k}{a_2 = k a_1, \quad b_2 = k b_1, \quad c_2 = k c_1}

    The two triangles have two equal angles, the right angle and the one of measure α\alpha; by the two-angle similarity criterion, they are similar, and their corresponding sides are proportional: there is a number k>0k > 0 such that each side of the second is kk times the corresponding side of the first.

  2. a2c2=ka1kc1=a1c1\frac{a_2}{c_2} = \frac{\resaltar{k a_1}}{\resaltar{k c_1}} = \resaltar{\frac{a_1}{c_1}}

    The quotient of the opposite side by the hypotenuse is the same in both triangles, because the factor kk cancels.

  3. b2c2=kb1kc1=b1c1\frac{b_2}{c_2} = \frac{\resaltar{k b_1}}{\resaltar{k c_1}} = \resaltar{\frac{b_1}{c_1}}

    The same happens with the quotient of the adjacent side by the hypotenuse.

  4. a2b2=ka1kb1=a1b1\frac{a_2}{b_2} = \frac{\resaltar{k a_1}}{\resaltar{k b_1}} = \resaltar{\frac{a_1}{b_1}}

    And with the quotient of the two legs. Therefore, the sine, the cosine and the tangent, and with them their reciprocals, take the same value in the two triangles: they depend only on α\alpha.

  5. c2=a2+b2→0<ac<1,0<bc<1c^2 = a^2 + b^2 \rightarrow 0 < \frac{a}{c} < 1, \quad 0 < \frac{b}{c} < 1

    By the Pythagorean theorem, the square of the hypotenuse is the sum of the squares of the legs, and exceeds each of them: the hypotenuse is greater than each leg, and the two quotients with the hypotenuse lie between 00 and 11.

The theorem allows the ratios of an angle to be computed with any right triangle containing it, choosing the most convenient one. In particular, the leg opposite one of the acute angles is the leg adjacent to the other, and from this follows a relation between the ratios of two angles that add up to π2\frac{\pi}{2}, called complementary.

ExampleThe ratios of the two acute angles

In a right triangle with legs 33 and 44, let α\alpha be the angle opposite the leg 33 and β\beta the one opposite the leg 44. Compute their ratios and compare them.

Proof

  1. c=32+42=25=5c = \sqrt{\resaltar{3^2 + 4^2}} = \sqrt{25} = \resaltar{5}

    By the Pythagorean theorem, the hypotenuse measures 55.

  2. sin⁡α=35,cos⁡α=45,tan⁡α=34\dato{a}{\sin \alpha = \frac{3}{5}, \quad \cos \alpha = \frac{4}{5}, \quad \tan \alpha = \frac{3}{4}}

    For α\alpha, the opposite side is 33 and the adjacent side is 44.

  3. sin⁡β=45,cos⁡β=35,tan⁡β=43\dato{b}{\sin \beta = \frac{4}{5}, \quad \cos \beta = \frac{3}{5}, \quad \tan \beta = \frac{4}{3}}

    For β\beta, the roles of the legs are exchanged: the opposite side is 44 and the adjacent side is 33.

  4. sin⁡β=cos⁡α,cos⁡β=sin⁡α,tan⁡β=1tan⁡α=cot⁡α\sin \beta = \cos \alpha, \quad \cos \beta = \sin \alpha, \quad \tan \beta = \frac{1}{\tan \alpha} = \cot \alpha

    Since α+β=π2\alpha + \beta = \frac{\pi}{2}, the sine of each angle is the cosine of the other, and the tangent of each is the cotangent of the other. The same happens in every right triangle, because the exchange of the legs does not depend on their measures.

This is where the names come from: the cosine is the sine of the complement, the cotangent is the tangent of the complement and the cosecant is the secant of the complement.