Define sine, cosine and tangent as side ratios that depend only on the acute angle, and explain why the sine of an angle equals the cosine of its complement

Learning goal: Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles; and explain and use the relationship between the sine and cosine of complementary angles.

Why a ratio can belong to an angle. Take any right triangle containing the acute angle θ whose opposite and adjacent legs are in the ratio 3 to 4; θ ≈ 36.87° to two decimal places. One such triangle has legs 3 and 4 and hypotenuse 5. Another has legs 6 and 8 and hypotenuse 10; another 1.5, 2 and 2.5; another 4.5, 6 and 7.5. All right triangles containing the angle θ are similar to each other by AA, since the right angle supplies the second pair of equal angles. Similar triangles have corresponding sides in one ratio, so the side lengths differ from triangle to triangle, but any ratio of two sides within a triangle is the same in every one of them: opposite over hypotenuse is 3/5 = 6/10 = 1.5/2.5 = 4.5/7.5 = 0.6 in all four. That number does not depend on which triangle was drawn. It depends only on the angle. In that sense side ratios in right triangles are properties of the angles, and it makes sense to give the ratios names as functions of the angle.

The definitions. In a right triangle, choose one acute angle and call it θ. The hypotenuse is the side opposite the right angle, always the longest side. The side opposite θ is the leg that does not touch θ; the side adjacent to θ is the leg that does. Then sine of θ is opposite ÷ hypotenuse; cosine of θ is adjacent ÷ hypotenuse; tangent of θ is opposite ÷ adjacent. These definitions apply to acute angles, between 0° and 90°, because only an acute angle can sit in a right triangle with the right angle. For this angle θ ≈ 36.87°, sin θ = 3/5 = 0.6, cos = 4/5 = 0.8 and tan = 3/4 = 0.75, in every right triangle containing that angle. Because the hypotenuse is the longest side, sine and cosine of an acute angle are always between 0 and 1; tangent can be any positive number.

The words "opposite" and "adjacent" are relative to the angle. In the 3-4-5 triangle, the other acute angle, 90° − θ ≈ 53.13°, has opposite leg 4 and adjacent leg 3. So sin(90° − θ) = 4/5 = 0.8, cos(90° − θ) = 3/5 = 0.6 and tan(90° − θ) = 4/3; with the rounded angle, sin 53.13° ≈ 0.8. Every ratio is computed from the point of view of the chosen angle. Mislabeling opposite and adjacent is the single most common error, and it is avoided by writing the angle first and marking the sides second.

Two exact families. The 45°-45°-90° triangle has legs 1 and 1 and hypotenuse √2, so sin 45° = cos 45° = 1/√2 = √2/2 ≈ 0.7071 and tan 45° = 1. The 30°-60°-90° triangle has sides 1, √3, 2 with 1 opposite the 30° angle, so sin 30° = 1/2, cos 30° = √3/2 ≈ 0.8660, tan 30° = 1/√3 = √3/3 ≈ 0.5774, and from the 60° vertex, sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3 ≈ 1.732. Those triangles are exact because they come from a square cut along a diagonal and an equilateral triangle cut along an altitude. For other angles a calculator supplies decimals; for instance sin 25° ≈ 0.4226.

Complementary angles. The two acute angles of a right triangle add to 90°, because the three angles add to 180° and one is 90°; so they are complementary. Call them θ and 90° − θ. The leg opposite θ is the leg adjacent to 90° − θ, and the hypotenuse is shared. Therefore sin θ = opposite-to-θ ÷ hypotenuse = adjacent-to-(90° − θ) ÷ hypotenuse = cos(90° − θ). In words: the sine of an angle equals the cosine of its complement, and the cosine of an angle equals the sine of its complement. That is the origin of the name cosine, "complement's sine". In the 3-4-5 triangle: sin θ = 0.6 = cos(90° − θ), and cos θ = 0.8 = sin(90° − θ), where θ ≈ 36.87° and 90° − θ ≈ 53.13°. With the exact families: sin 30° = cos 60° = 1/2 and sin 45° = cos 45°, since 45° is its own complement. With a calculator: sin 25° ≈ 0.4226 and cos 65° ≈ 0.4226.

Using the relationship. If cos θ = sin 20° and θ is acute, then θ = 70°, because cos 70° = sin(90° − 70°) = sin 20°, and cosine takes each value only once on the acute angles. If sin(2x)° = cos(x + 15)° with both angles acute, then 2x and x + 15 are complementary: 2x + x + 15 = 90, so x = 25, giving sin 50° = cos 40°. The relationship also halves the work of building a table: once the sines of 0° to 45° are known, the cosines of 45° to 90° are the same numbers in reverse order.

The misconception. Students sometimes believe a bigger triangle has a bigger sine. Doubling every side of a right triangle doubles the opposite side and doubles the hypotenuse, so their ratio is unchanged; the sine belongs to the angle, not to the triangle. Conversely, a bigger angle does have a bigger sine, because as θ grows toward 90° the opposite leg grows toward the length of the hypotenuse; and cosine shrinks at the same time, which is the complement relationship seen from another side.

Worked, ratios from a triangle: A right triangle has legs 5 and 12, so its hypotenuse is 13. Let θ be the angle opposite the leg 5. Then sin θ = 5/13 ≈ 0.3846, cos θ = 12/13 ≈ 0.9231, tan θ = 5/12 ≈ 0.4167. For the other acute angle, φ = 90° − θ, the roles swap: sin φ = 12/13, cos φ = 5/13, tan φ = 12/5 = 2.4. Notice sin θ = cos φ and cos θ = sin φ, as complementary angles require.

Worked, why the size of the triangle does not matter: Triangle 1 has legs 3, 4 and hypotenuse 5; triangle 2 has legs 9, 12 and hypotenuse 15. Both have a right angle and both have the angle whose tangent is 3/4, so they are similar by AA with scale factor 3. For that angle, sin = 3/5 in triangle 1 and 9/15 = 3/5 in triangle 2; cos = 4/5 and 12/15 = 4/5. The sides tripled, the ratios did not; the ratios are properties of the angle.

Worked, exact values: In a 30-60-90 triangle with sides 1, √3, 2, the side 1 is opposite 30°. So sin 30° = 1/2, cos 30° = √3/2, and sin 60° = √3/2, cos 60° = 1/2. The complement relationship sin 30° = cos 60° and cos 30° = sin 60° is visible directly: the leg opposite 30° is the leg adjacent to 60°.

Worked, solving with complements: Find x when sin(3x)° = cos(x + 10)° and both 3x and x + 10 are acute angles. For acute angles, the sine of one equals the cosine of the other exactly when they are complementary, so 3x and x + 10 must be complementary: 3x + x + 10 = 90, so 4x = 80 and x = 20. Check: sin 60° = √3/2 and cos 30° = √3/2. Both 60° and 30° are acute, so the solution is valid. The condition matters: if only x had to be acute, x = 50 would also work, since sin 150° = cos 60° = 1/2, but 150° is not acute.

Recording task: Draw four right triangles with legs 3 and 4, 6 and 8, 1.5 and 2, and 4.5 and 6 (in centimeters). Measure each hypotenuse and record it beside the computed values 5, 10, 2.5 and 7.5. For the angle opposite the shorter leg, compute opposite/hypotenuse, adjacent/hypotenuse and opposite/adjacent in each triangle and record the table; every column should read 0.6, 0.8 and 0.75. Measure the angle with a protractor in each triangle (expect about 37°) and record it. Then, for the other acute angle (about 53°), record the same three ratios and mark which entries are equal to entries in the first table, with the reason "complementary angles".

Independent performance: Write an explanation, using the AA criterion, of why the ratio opposite/hypotenuse for a 40° angle is the same in every right triangle that contains a 40° angle, and why that justifies naming the ratio sin 40°. Then, for a right triangle with legs 7 and 24, compute all six ratios (sine, cosine and tangent of both acute angles) as fractions and as decimals, and pair up the equal ones with the complement relationship. Finally solve sin(2x + 5)° = cos(x − 5)°, given that 2x + 5 and x − 5 are both acute angles, and check the solution numerically. Success criteria: the explanation names the right angle as the second pair of equal angles; the hypotenuse is computed correctly as 25; every fraction is reduced and paired correctly; the equation is solved by making the two angles complementary; and both angles in the check are shown to be acute.