Use special right triangles to find exact values of sine, cosine and tangent at π/6, π/4 and π/3

Learning goal: Use the 45-45-90 and 30-60-90 special triangles to determine geometrically the exact values of sine, cosine and tangent for π/3, π/4 and π/6, place those values on the unit circle, and begin using the unit circle to express the values at π − x, π + x and 2π − x in terms of the values at x.

Scope note. F.TF.3 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

Why exact values matter. A calculator gives sin(π/3) ≈ 0.8660, and for many purposes that is enough. But 0.8660 is a rounded number, and the true value is √3/2, an exact number that comes from the geometry of the equilateral triangle. Knowing the exact values at the three special angles lets you solve equations exactly, check a calculator against something you can trust, and recognize geometric structure. There are only three angles to learn, and each comes from one triangle you can draw and reason about.

The 45-45-90 triangle and π/4. Take a square with side 1 and cut it along a diagonal. Each half is a right triangle with two legs of length 1 and two 45° angles, and 45° = π/4. By the Pythagorean theorem the hypotenuse is √(1² + 1²) = √2. So in this triangle sin(π/4) = opposite/hypotenuse = 1/√2, cos(π/4) = adjacent/hypotenuse = 1/√2 and tan(π/4) = opposite/adjacent = 1/1 = 1. The fraction 1/√2 is usually written √2/2: multiply top and bottom by √2 to get √2/2 ≈ 0.7071. Sine and cosine are equal at π/4 because the two legs are equal, and the tangent is 1 for the same reason.

The 30-60-90 triangle, π/6 and π/3. Take an equilateral triangle with side 2 and cut it in half along an altitude. Each half is a right triangle whose hypotenuse is the original side, 2, and whose short leg is half of a side, 1. The altitude is the long leg, and by the Pythagorean theorem it is √(2² − 1²) = √3. The angles are 30° = π/6, opposite the short leg, and 60° = π/3, opposite the long leg. Reading ratios from this triangle: sin(π/6) = 1/2, cos(π/6) = √3/2, tan(π/6) = 1/√3 = √3/3 ≈ 0.5774; and sin(π/3) = √3/2 ≈ 0.8660, cos(π/3) = 1/2, tan(π/3) = √3/1 = √3 ≈ 1.7321. The two angles swap sine and cosine, because the side opposite π/6 is the side adjacent to π/3, and their tangents are reciprocals of each other.

A fully worked model. Derive the values at π/6 from scratch and place them on the unit circle. Start with an equilateral triangle of side 2 and drop the altitude from one vertex to the opposite side; it bisects that side into two pieces of length 1 and bisects the 60° vertex angle into two 30° angles. In one half, the hypotenuse is 2, the leg adjacent to the 30° angle is the altitude √3, and the leg opposite it is 1. So cos(π/6) = √3/2 and sin(π/6) = 1/2. On the unit circle, P(π/6) is therefore the point (√3/2, 1/2) ≈ (0.866, 0.5). Check that it lies on the circle: (√3/2)² + (1/2)² = 3/4 + 1/4 = 1. The check passes, and the point is in quadrant I where both coordinates are positive, as they should be for an angle between 0 and π/2.

Scaling the triangle to fit the unit circle. The triangles above have hypotenuses √2 and 2, not 1. That does not matter, because the ratios do not depend on the size of the triangle. If you prefer a triangle that sits inside the unit circle, divide every side by the hypotenuse: the 45-45-90 triangle becomes legs √2/2, √2/2 and hypotenuse 1, and the 30-60-90 triangle becomes legs 1/2 and √3/2 with hypotenuse 1. Then the legs are the coordinates of the point on the unit circle, which is what the previous lesson's definition uses.

The pattern in the table. Written as square roots, the sines of 0, π/6, π/4, π/3 and π/2 are √0/2, √1/2, √2/2, √3/2 and √4/2, which equal 0, 1/2, √2/2, √3/2 and 1. The cosines run the same list backwards. This is a memory aid, not a proof; the proof is the two triangles. The tangents are the quotients: 0, √3/3, 1, √3 and undefined.

Beyond the first quadrant. The unit circle turns three exact values into infinitely many. Every angle that is not on an axis has a reference angle, the acute angle between its terminal side and the x-axis, and the point for the angle has the same coordinates as the point for the reference angle up to sign. In particular, when x is between 0 and π/2, the angle π − x has reference angle x and lands in quadrant II, so cos(π − x) = −cos x and sin(π − x) = sin x; the angle π + x lands in quadrant III, so cos(π + x) = −cos x and sin(π + x) = −sin x; and 2π − x lands in quadrant IV, so cos(2π − x) = cos x and sin(2π − x) = −sin x. For example sin(5π/6) = sin(π − π/6) = sin(π/6) = 1/2, and cos(4π/3) = cos(π + π/3) = −cos(π/3) = −1/2. These relationships hold for every real number x, not only acute ones, and tangent follows by dividing sine by cosine: tan(π − x) = −tan x, tan(π + x) = tan x and tan(2π − x) = −tan x wherever tan x is defined; for example tan(5π/6) = −tan(π/6) = −√3/3. The next lesson proves all three for every real x from the symmetry of the circle, together with periodicity.

The misconception. A common error is to write sin(π/6) = √3/2 and cos(π/6) = 1/2, swapping the values. The cure is to draw the triangle every time until the picture is automatic: π/6 is the small angle, so the side opposite it is the short leg 1, and sin(π/6) = 1/2 is the smaller of the two values. Sine grows as the angle grows from 0 to π/2, so sin(π/6) < sin(π/4) < sin(π/3), that is 0.5 < 0.707 < 0.866, and cosine falls in the same range.

Worked, deriving π/4 exactly: In an isosceles right triangle with legs 1, the hypotenuse is √(1 + 1) = √2. So sin(π/4) = 1/√2. Rationalize the denominator: multiply numerator and denominator by √2 to get √2/2. Then cos(π/4) = √2/2 as well, and tan(π/4) = 1. Check on the unit circle: (√2/2)² + (√2/2)² = 2/4 + 2/4 = 1, so P(π/4) = (√2/2, √2/2) is on the circle, and the point lies on the line y = x, as it must for the angle that bisects the first quadrant.

Worked, using an exact value in geometry: A ramp rises at an angle of π/6 to the ground and its sloping surface is 5 m long. How high does it rise, exactly and to the nearest centimeter? The height is the side opposite π/6, so height = 5 sin(π/6) = 5 × 1/2 = 2.5 m exactly. The horizontal run is 5 cos(π/6) = 5√3/2 ≈ 4.33 m. Check with the Pythagorean theorem using the exact run: 2.5² + (5√3/2)² = 6.25 + 75/4 = 6.25 + 18.75 = 25 = 5².

Worked, an exact value at a reference angle: Find tan(2π/3) exactly. The angle 2π/3 = π − π/3 lands in quadrant II with reference angle π/3. Cosine is negative there and sine positive: cos(2π/3) = −1/2 and sin(2π/3) = √3/2. So tan(2π/3) = (√3/2)/(−1/2) = −√3 ≈ −1.732. Calculator check in radian mode: tan(2.0944) ≈ −1.732.

Worked, from a value back to an angle: For which angle x between 0 and π/2 is cos x = √3/2? From the 30-60-90 triangle, cos(π/6) = √3/2, and cosine is decreasing on [0, π/2], so π/6 is the only such angle in that interval. On the unit circle there is a second point with x-coordinate √3/2, namely (√3/2, −1/2) at angle 11π/6 = 2π − π/6, which is outside the given interval.

Representation task: Construct the two special triangles with ruler and compass or on grid paper: a square of side 4 cm cut along its diagonal, and an equilateral triangle of side 4 cm cut along an altitude. Measure the diagonal and the altitude and compare with 4√2 ≈ 5.66 cm and 2√3 ≈ 3.46 cm. Then make a table with columns angle in radians, angle in degrees, sin, cos, tan, each as an exact value with radicals and as a decimal to four places, for π/6, π/4 and π/3. Add a row for 0 and for π/2 using the unit circle points (1, 0) and (0, 1).

Independent performance: Write up a derivation of all nine values (sine, cosine and tangent at π/6, π/4 and π/3) from the two triangles, showing the Pythagorean step for each hypotenuse or leg, rationalizing every denominator, and checking each pair (cos, sin) against x² + y² = 1. Then use the values to solve two problems you write yourself: one that finds a length in a right triangle with a π/3 or π/6 angle, and one that finds the coordinates of the point on the unit circle at 5π/6, 5π/4 or 5π/3, with a sentence explaining the sign of each coordinate. Success criteria: all nine values are correct; every radical is simplified with no radical in a denominator; the two applications are correct and carry units where lengths are involved; and the coordinate answer names the quadrant and the reference angle.