The Four Quadrants
The four quadrants extend trigonometry to any angle from \(0^\circ\) to \(360^\circ\) and beyond, using the unit circle. The ASTC rule shows which ratios are positive in each quadrant, and each angle has a related acute angle, so \(\cos 240^\circ=-\cos 60^\circ\). This is NSW Year 10 Advanced maths (NESA outcome MA5‑TRG‑P‑02).
You will decide where each ratio is positive or negative, find exact values for angles such as \(315^\circ\), \(-60^\circ\) and \(480^\circ\), and find the coordinates of a point on the unit circle. You will solve equations such as \(2\cos\theta-\sqrt{3}=0\) and \(\tan^2\theta=3\) for every angle from \(0^\circ\) to \(360^\circ\), and use a point on the terminal arm, or one ratio and a quadrant, to find the other ratios. It is part of Year 10 Advanced Trigonometry.
Theory
Any angle, even one over \(360^\circ\) or a negative one, has a sine, cosine and tangent. Measure it anticlockwise from the positive \(x\)-axis, find its quadrant and its related acute angle, then take the sign from the ASTC rule. This Year 10 Advanced maths page covers the unit circle, exact values in all four quadrants and solving equations from \(0^\circ\) to \(360^\circ\). It leads straight into Mathematics Advanced in Years 11 and 12.
On the unit circle (radius \(1\), centre the origin), the point at angle \(\theta\), measured anticlockwise from the positive \(x\)-axis, is \(P(\cos\theta, \sin\theta)\). This defines sine and cosine for every angle, and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\).
The axes split the plane into four quadrants. The signs of \(x\) and \(y\) give the ASTC rule: All positive in quadrant 1, Sine in 2, Tangent in 3, Cosine in 4.
The related acute angle is the angle between the arm \(OP\) and the nearest part of the \(x\)-axis. A ratio of \(\theta\) equals the same ratio of the related angle, with the sign from ASTC.
A negative angle is measured clockwise. Adding or subtracting \(360^\circ\) does not change the direction, so it does not change any ratio: \(\sin 400^\circ = \sin 40^\circ\). For any point \((x, y)\) on the arm, at distance \(r\) from the origin: \(\sin\theta = \dfrac{y}{r}\), \(\cos\theta = \dfrac{x}{r}\), \(\tan\theta = \dfrac{y}{x}\).
| Quadrant | Angle \(\theta\) | Related angle | Positive |
|---|---|---|---|
| 1 | \(0^\circ\)–\(90^\circ\) | \(\theta\) | all |
| 2 | \(90^\circ\)–\(180^\circ\) | \(180^\circ - \theta\) | \(\sin\) |
| 3 | \(180^\circ\)–\(270^\circ\) | \(\theta - 180^\circ\) | \(\tan\) |
| 4 | \(270^\circ\)–\(360^\circ\) | \(360^\circ - \theta\) | \(\cos\) |
How to find a ratio of any angle
- Bring the angle into \(0^\circ\)–\(360^\circ\) by adding or subtracting \(360^\circ\).
- Find the quadrant and the related acute angle (table above).
- Write the ratio of the related angle, with the sign from ASTC.
How to solve an equation for \(0^\circ \le \theta \le 360^\circ\)
- Isolate the ratio, e.g. \(2\sin\theta + \sqrt{3} = 0\) gives \(\sin\theta = -\dfrac{\sqrt{3}}{2}\). For a square such as \(\sin^2\theta = k\), take both square roots.
- Find the related angle from the positive value, by exact values or \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\).
- Choose the two quadrants where the ratio has the right sign, and place the related angle in each: \(\alpha\), \(180^\circ - \alpha\), \(180^\circ + \alpha\) or \(360^\circ - \alpha\).
Two conditions
- If you are told, for example, \(\sin\theta > 0\) and \(\tan\theta < 0\), list the quadrants for each condition and take the one they share (here quadrant 2).
- Then draw the related-angle triangle, find the third side, and attach the signs for that quadrant (Example 3).
(i) \(\sin 240^\circ\)
(ii) \(\cos 315^\circ\)
(iii) \(\tan(-150^\circ)\)
(iv) \(\sin 480^\circ\)
For each angle: find the quadrant, the related acute angle, then the sign from ASTC.
(i) Quadrant 3, related angle \(240^\circ - 180^\circ = 60^\circ\); sine is negative there.
| \(\sin 240^\circ\) | \(=\) | \(-\sin 60^\circ = \mathbf{-\dfrac{\sqrt{3}}{2}}\) |
(ii) Quadrant 4, related angle \(360^\circ - 315^\circ = 45^\circ\); cosine is positive there.
| \(\cos 315^\circ\) | \(=\) | \(\cos 45^\circ = \mathbf{\dfrac{1}{\sqrt{2}}}\) |
(iii) Add \(360^\circ\): \(-150^\circ\) points the same way as \(210^\circ\). That is quadrant 3, where tangent is positive.
| \(\tan(-150^\circ)\) | \(=\) | \(\tan 210^\circ\) |
| \(\) | \(=\) | \(\tan 30^\circ = \mathbf{\dfrac{1}{\sqrt{3}}}\) |
(iv) Subtract \(360^\circ\) to get \(120^\circ\), in quadrant 2, where sine is positive.
| \(\sin 480^\circ\) | \(=\) | \(\sin 120^\circ = \sin 60^\circ = \mathbf{\dfrac{\sqrt{3}}{2}}\) |
(i) \(2\cos\theta - 1 = 0\)
(ii) \(\sin\theta = -0.7\), to the nearest degree
(iii) \(\tan^2\theta = 1\)
(i) \(\cos\theta = \dfrac{1}{2}\), related angle \(60^\circ\). Cosine is positive in quadrants 1 and 4.
| \(\theta\) | \(=\) | \(\mathbf{60^\circ}\text{ or } 360^\circ - 60^\circ = \mathbf{300^\circ}\) |
(ii) Find the related angle from the positive value. Sine is negative in quadrants 3 and 4.
| \(\alpha\) | \(=\) | \(\sin^{-1}(0.7) = 44.427\ldots^\circ\) |
| \(\theta\) | \(=\) | \(180^\circ + \alpha = 224.427\ldots^\circ\) |
| \(\text{or } \theta\) | \(=\) | \(360^\circ - \alpha = 315.572\ldots^\circ\) |
| \(\theta\) | \(\approx\) | \(\mathbf{224^\circ}\text{ or }\mathbf{316^\circ}\) |
(iii) \(\tan\theta = \pm 1\), related angle \(45^\circ\). Both signs count, so there is one answer in each quadrant.
| \(\theta\) | \(=\) | \(\mathbf{45^\circ},\ \mathbf{135^\circ},\ \mathbf{225^\circ},\ \mathbf{315^\circ}\) |
Draw the triangle for the related angle \(\alpha\) with adjacent \(5\) and hypotenuse \(13\).
| \(\text{opposite}\) | \(=\) | \(\sqrt{13^2 - 5^2} = \sqrt{144} = 12\) |
\(\theta\) is in quadrant 4, where only cosine is positive.
| \(\sin\theta\) | \(=\) | \(\mathbf{-\dfrac{12}{13}}\) |
| \(\tan\theta\) | \(=\) | \(\dfrac{\sin\theta}{\cos\theta} = \mathbf{-\dfrac{12}{5}}\) |
(i) the exact values of \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\)
(ii) \(\theta\), correct to the nearest degree.
(i) Find \(r\), the distance from the origin, then use \(\sin\theta = \dfrac{y}{r}\), \(\cos\theta = \dfrac{x}{r}\), \(\tan\theta = \dfrac{y}{x}\). The signs come out automatically.
| \(r\) | \(=\) | \(\sqrt{(-5)^2 + 12^2} = 13\) |
| \(\sin\theta\) | \(=\) | \(\mathbf{\dfrac{12}{13}}\) |
| \(\cos\theta\) | \(=\) | \(\mathbf{-\dfrac{5}{13}}\) |
| \(\tan\theta\) | \(=\) | \(\dfrac{12}{-5} = \mathbf{-\dfrac{12}{5}}\) |
(ii) Quadrant 2. The related angle has \(\tan\alpha = \dfrac{12}{5}\).
| \(\alpha\) | \(=\) | \(67.380\ldots^\circ\) |
| \(\theta\) | \(=\) | \(180^\circ - \alpha = 112.619\ldots^\circ \approx \mathbf{113^\circ}\) |
A calculator's \(\tan^{-1}\left(-\dfrac{12}{5}\right)\) gives \(-67.4^\circ\), which is the wrong quadrant. Always work from the related angle.
Common pitfalls
Frequently asked questions
What does ASTC mean in trigonometry?
ASTC tells you which ratios are positive in each quadrant, going anticlockwise from the first: All in quadrant 1, Sine in quadrant 2, Tangent in quadrant 3 and Cosine in quadrant 4. A common way to remember it is All Stations To Central.
How do you find the related angle?
It is the acute angle between the arm of the angle and the x-axis. In quadrant 2 it is 180 minus theta, in quadrant 3 it is theta minus 180, and in quadrant 4 it is 360 minus theta.
Why does a trig equation have two answers between 0 and 360 degrees?
Each ratio is positive in two quadrants and negative in the other two. So a value such as \(\sin\theta = \dfrac{1}{2}\) is reached once in each of two quadrants, here at \(30^\circ\) and \(150^\circ\).
How do you find sin of a negative angle?
A negative angle is measured clockwise. Add \(360^\circ\) to get the same direction as a positive angle, or use \(\sin(-\theta) = -\sin\theta\).
What is the unit circle definition of sine and cosine?
On a circle of radius 1 centred at the origin, the point at angle theta from the positive x-axis has coordinates cos theta, sin theta. Cosine is the x-coordinate and sine is the y-coordinate.