This note covers the unit circle definitions of sine, cosine and tangent for any angle, the exact values you are expected to know, and the ambiguous case of the sine rule, where two different triangles fit the same information. These ideas extend trigonometry beyond right-angled triangles and acute angles.
Key ideas
- The unit circle has radius 1 and its centre at the origin. Take a point \(P\) on it, where the line from the origin to \(P\) makes an angle \(\theta\) measured anticlockwise from the positive \(x\)-axis. Then \(P = (\cos\theta, \sin\theta)\).
- So \(\cos\theta\) is the \(x\)-coordinate and \(\sin\theta\) is the \(y\)-coordinate. This works for any angle, including obtuse, reflex and negative angles.
- \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\). It is the gradient of the line \(OP\), and it is undefined when \(\cos\theta = 0\).
- Because the point is on a circle of radius 1, \(\sin^2\theta + \cos^2\theta = 1\).
- The quadrants decide the signs. In the first quadrant all three are positive. In the second only \(\sin\) is positive. In the third only \(\tan\) is positive. In the fourth only \(\cos\) is positive.
- Symmetry gives related angles: \(\sin(180^\circ - \theta) = \sin\theta\), \(\cos(180^\circ - \theta) = -\cos\theta\), \(\sin(-\theta) = -\sin\theta\), \(\cos(-\theta) = \cos\theta\). In radians replace \(180^\circ\) with \(\pi\).
- To find a value for a big angle, use the reference angle, the acute angle between the line \(OP\) and the \(x\)-axis. Then give the sign from the quadrant.
Ambiguous case. If you know two sides and an angle that is not between them (side \(a\), side \(b\) and angle \(A\)), the sine rule gives \(\sin B\). Since \(\sin B = \sin(180^\circ - B)\), there can be two possible angles \(B\): an acute one and an obtuse one. The second one is valid only if it still leaves room for the angles to sum to \(180^\circ\).
- If \(a \geq b\), there is only one triangle (the obtuse \(B\) would not fit).
- If \(b\sin A \lt a \lt b\), there are two triangles.
- If \(a = b\sin A\), there is one right-angled triangle.
- If \(a \lt b\sin A\), there is no triangle.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(P = (\cos\theta, \sin\theta)\) | Coordinates of a point on the unit circle at angle \(\theta\). |
| \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) | Gradient of the radius. Undefined at \(90^\circ\) and \(270^\circ\). |
| \(\sin 30^\circ = \dfrac{1}{2},\ \cos 30^\circ = \dfrac{\sqrt{3}}{2},\ \tan 30^\circ = \dfrac{1}{\sqrt{3}}\) | Exact values for \(\frac{\pi}{6}\). |
| \(\sin 45^\circ = \cos 45^\circ = \dfrac{\sqrt{2}}{2},\ \tan 45^\circ = 1\) | Exact values for \(\frac{\pi}{4}\). |
| \(\sin 60^\circ = \dfrac{\sqrt{3}}{2},\ \cos 60^\circ = \dfrac{1}{2},\ \tan 60^\circ = \sqrt{3}\) | Exact values for \(\frac{\pi}{3}\). |
| \(\sin 0 = 0,\ \sin\frac{\pi}{2} = 1,\ \cos 0 = 1,\ \cos\frac{\pi}{2} = 0\) | Axis angles: \(0\), \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\). |
| \(\dfrac{\sin B}{b} = \dfrac{\sin A}{a}\) | Sine rule used to find \(B\), which may have two solutions. |
Worked examples
Example 1 Find the exact values of (a) \(\sin\dfrac{5\pi}{6}\), (b) \(\cos 210^\circ\), (c) \(\tan\dfrac{7\pi}{4}\).
(a) \(\frac{5\pi}{6} = \pi - \frac{\pi}{6}\) is in the second quadrant, where sine is positive. The reference angle is \(\frac{\pi}{6}\), so \(\sin\frac{5\pi}{6} = \sin\frac{\pi}{6} = \frac{1}{2}\).
(b) \(210^\circ = 180^\circ + 30^\circ\) is in the third quadrant, where cosine is negative. The reference angle is \(30^\circ\), so \(\cos 210^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}\).
(c) \(\frac{7\pi}{4} = 2\pi - \frac{\pi}{4}\) is in the fourth quadrant, where tangent is negative. The reference angle is \(\frac{\pi}{4}\), so \(\tan\frac{7\pi}{4} = -\tan\frac{\pi}{4} = -1\).
Answer: (a) \(\frac{1}{2}\); (b) \(-\frac{\sqrt{3}}{2}\); (c) \(-1\).
Example 2 The angle \(\theta\) is obtuse and \(\sin\theta = \dfrac{5}{13}\). Find \(\cos\theta\) and \(\tan\theta\) exactly.
Use \(\sin^2\theta + \cos^2\theta = 1\): \(\cos^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}\).
So \(\cos\theta = \pm\dfrac{12}{13}\). An obtuse angle is in the second quadrant, where cosine is negative, so \(\cos\theta = -\dfrac{12}{13}\).
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{5/13}{-12/13} = -\dfrac{5}{12}\).
Answer: \(\cos\theta = -\frac{12}{13}\), \(\tan\theta = -\frac{5}{12}\).
Example 3 In triangle \(ABC\), \(A = 30^\circ\), \(a = 7\) cm and \(b = 10\) cm. Find all possible triangles, giving angle \(B\) and side \(c\) in each.
Here \(b\sin A = 10 \sin 30^\circ = 5\), and \(5 \lt 7 \lt 10\), so two triangles are possible.
\(\sin B = \dfrac{10 \sin 30^\circ}{7} = 0.7143\). The acute solution is \(B = \sin^{-1}(0.7143) \approx 45.6^\circ\). The obtuse solution is \(B = 180^\circ - 45.6^\circ = 134.4^\circ\).
Triangle 1: \(B = 45.6^\circ\), so \(C = 180^\circ - 30^\circ - 45.6^\circ = 104.4^\circ\). Then \(c = \dfrac{7 \sin 104.4^\circ}{\sin 30^\circ} \approx 13.6\) cm.
Triangle 2: \(B = 134.4^\circ\), so \(C = 180^\circ - 30^\circ - 134.4^\circ = 15.6^\circ\). Then \(c = \dfrac{7 \sin 15.6^\circ}{\sin 30^\circ} \approx 3.76\) cm.
Answer: \(B \approx 45.6^\circ,\ c \approx 13.6\) cm, or \(B \approx 134.4^\circ,\ c \approx 3.76\) cm.
Common errors
- Giving only the acute angle from \(\sin^{-1}\) and missing the second solution when the information is side-side-angle.
- Accepting an obtuse second angle without checking that \(A + B\) is less than \(180^\circ\), so that the third angle is still positive.
- Getting the sign wrong for angles outside the first quadrant. Name the quadrant before writing the sign.
- Writing \(\tan 90^\circ = \infty\). It is undefined.
- Mixing up the exact values of \(\sin 30^\circ\) and \(\sin 60^\circ\), or using \(\frac{\sqrt{3}}{3}\) and \(\frac{1}{\sqrt{3}}\) inconsistently (they are equal).
Exam tip: For exact values, draw a quick unit circle sketch, shade the quadrant and write the sign before you write the number. This catches most sign slips.
Check yourself
- Find the exact values of \(\sin 300^\circ\) and \(\cos\dfrac{2\pi}{3}\).
Answer
\(300^\circ\) is in the fourth quadrant with reference angle \(60^\circ\), so \(\sin 300^\circ = -\frac{\sqrt{3}}{2}\). \(\frac{2\pi}{3} = \pi - \frac{\pi}{3}\) is in the second quadrant, so \(\cos\frac{2\pi}{3} = -\frac{1}{2}\). - Given \(\cos\theta = -\dfrac{3}{5}\) and \(\theta\) in the third quadrant, find \(\sin\theta\) and \(\tan\theta\).
Answer
\(\sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}\). In the third quadrant sine is negative, so \(\sin\theta = -\frac{4}{5}\). \(\tan\theta = \frac{-4/5}{-3/5} = \frac{4}{3}\). - In triangle \(ABC\), \(A = 35^\circ\), \(a = 6\) cm and \(b = 9\) cm. Find both possible values of \(B\).
Answer
\(b\sin A = 9 \sin 35^\circ \approx 5.16 \lt 6 \lt 9\), so there are two triangles. \(\sin B = \frac{9\sin 35^\circ}{6} \approx 0.8604\), so \(B \approx 59.4^\circ\) or \(B = 180^\circ - 59.4^\circ \approx 120.6^\circ\). Both leave a positive third angle (\(85.6^\circ\) or \(24.4^\circ\)). - In triangle \(ABC\), \(A = 30^\circ\), \(a = 3\) cm and \(b = 10\) cm. How many triangles are possible?
Answer
\(b\sin A = 10\sin 30^\circ = 5\), which is greater than \(a = 3\). The side \(a\) is too short to reach the base line, so no triangle exists. Check: \(\sin B = \frac{10 \times 0.5}{3} = 1.67\), which is impossible.