Trigonometric Functions
Exact Values, Reference Angles and Quadrant Signs
The two special triangles, the ASTC sign rule, and the reference angle method that reduces any angle to an acute one.
What this page covers
- Derive the exact values for 30°, 45° and 60°
- State which functions are positive in each quadrant
- Find the reference angle for any angle
- Combine the two to evaluate a function at any special angle
The two special triangles
Three angles have exact values expressible in surds, and both come from elementary geometry rather than memory.
The 45° triangle
Cut a unit square along its diagonal. The two legs are 1 and 1, and the hypotenuse is √2 by Pythagoras. The acute angles are both 45°.
Week 8, page 5 gives tanπ4 = 1√2 in the radian column, while the degree column on the same page correctly gives tan 45° = 1. Since π4 and 45° are the same angle, the radian entry is wrong. Tangent is opposite over adjacent, which here is 11 = 1.
The 30°–60° triangle
Cut an equilateral triangle of side 2 down its axis of symmetry. The result has hypotenuse 2, short leg 1, and long leg √22 - 12 = √3. The angles are 30° and 60°.
The values swap between 30° and 60°, because the two angles are complementary and what is opposite one is adjacent to the other.
The exact-value table
| Degrees | Radians | sin | cos | tan |
|---|---|---|---|---|
| 0° | 0 | 0 | 1 | 0 |
| 30° | π6 | 12 | √32 | 1√3 |
| 45° | π4 | 1√2 | 1√2 | 1 |
| 60° | π3 | √32 | 12 | √3 |
| 90° | π2 | 1 | 0 | undefined |
| 180° | π | 0 | -1 | 0 |
| 270° | 3pi2 | -1 | 0 | undefined |
Written as √02, √12, √22, √32, √42, the sine values across 0, 30, 45, 60, 90 degrees follow an obvious progression, and the cosine values are the same list reversed. Written this way the table is far easier to reconstruct than to memorise.
The source computes sinπ4 from the unit circle instead, which is a useful cross-check: at t = π4 the coordinates are equal, so x2 + x2 = 1, giving x = 1√2 ≈ 0.7071.
Quadrant signs: ASTC
Which functions are positive depends only on the signs of x and y in the quadrant concerned.
| Quadrant | x | y | sin | cos | tan | Positive |
|---|---|---|---|---|---|---|
| I (0–π2) | + | + | + | + | + | All |
| II (π2–π) | − | + | + | − | − | Sine |
| III (π–3pi2) | − | − | − | − | + | Tangent |
| IV (3pi2–2pi) | + | − | − | + | − | Cosine |
The source draws the quadrant diagram with A, S, T, C marked, reading anticlockwise from the first quadrant. Tangent is positive in the third because both coordinates are negative and the ratio yx is positive.
Secant has the sign of cosine, cosecant the sign of sine, and cotangent the sign of tangent, because a reciprocal never changes sign.
Reference angles
The acute angle between the terminal side and the horizontal axis. The source calls it the first angle and builds the whole solving method on it.
| Quadrant | Reference angle α | In degrees |
|---|---|---|
| I | θ | θ |
| II | π - θ | 180° - θ |
| III | θ - π | θ - 180° |
| IV | 2pi - θ | 360° - θ |
The value of any trigonometric function at θ equals its value at the reference angle α, with a sign supplied by ASTC.
Worked example — cos2pi3
- Locate the quadrant. 2pi3 = 120°, which is in quadrant II.
- Find the reference angle. π - 2pi3 = π3.
- Get the magnitude. cosπ3 = 12.
- Apply ASTC. In quadrant II only sine is positive, so cosine is negative.
A correction the source makes against itself
Week 8, page 4 solves cos θ = 12 on -π ≤ θ ≤ π and gives θ = ±2pi3. That is wrong: cos2pi3 = -12, as computed above. The correct answer is θ = ±π3.
The source itself has it right elsewhere. Week 8, page 8 solves the same equation and writes: reference angle π3, because cosπ3 = 12, therefore θ = π3 or -π3. Four pages apart, the notes disagree with themselves, and the later version is correct.
Cosine is positive in quadrants I and IV, so a positive cosine cannot have a solution in quadrant II. Checking the sign against ASTC before writing the answer catches it immediately.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| Measuring the reference angle to the vertical | Always to the horizontal | Otherwise sine and cosine swap |
| tanπ4 = 1√2 | = 1 | Opposite over adjacent is 11 |
| cos2pi3 = 12 | = -12 | Quadrant II: cosine is negative |
| Forgetting the sign entirely | Apply ASTC | The reference angle gives magnitude only |
| Confusing 30° and 60° values | sin 30° = 12 | The smaller angle has the smaller sine |
| Using the exact values in the wrong angle mode | π6 and 30° are the same angle | Check the mode |
Frequently asked questions
Do I have to memorise the exact values?
It is easier to remember the two triangles. A square cut along its diagonal gives the 45° values; an equilateral triangle cut in half gives the 30° and 60° ones. Both can be redrawn in seconds.
What is a reference angle?
The acute angle between the terminal side and the horizontal axis. Every trigonometric value at any angle equals the value at its reference angle, up to a sign fixed by the quadrant.
How do I remember ASTC?
All Stations To Central, or All Silly Tom Cats. Reading anticlockwise from the first quadrant: All positive, Sine only, Tangent only, Cosine only.
Why is the reference angle measured to the horizontal?
Because sine and cosine are defined by the coordinates, and the acute angle to the x-axis produces the same coordinate magnitudes in every quadrant. Measuring to the vertical would swap sine and cosine.
Source. Handwritten teaching notes, Week 8 pages 4-5 and Week 9 page 1, with the typed supplementary notes on trigonometry.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.