Two small triangles give you the exact sine, cosine and tangent of \(30^\circ\), \(45^\circ\) and \(60^\circ\); symmetry extends them to angles of any size. Two identities, \(\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}\) and \(\sin^2\theta + \cos^2\theta \equiv 1\), then do the rest.
Learn \(\sin 30^\circ = \tfrac{1}{2}\), \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\), \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\) (cosine runs the same list backwards) and get every related angle from its sign in that quadrant. For any \(\theta\): \(\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}\) and \(\sin^2\theta + \cos^2\theta \equiv 1\).
Fig. 1: half of an equilateral triangle of side 2 gives the 30° and 60° ratios; half of a unit square gives the 45° ratios. The missing sides come from Pythagoras
Section 01
The special triangles.
In the lab, step the angle through \(30^\circ\), \(45^\circ\) and \(60^\circ\) and compare the exact values it shows with the sides of the triangles in Fig. 1. You should be able to rebuild the whole table from the triangles in a few seconds, so you never need to memorise it blind.
\(\theta\)
\(0\)
\(30^\circ = \frac{\pi}{6}\)
\(45^\circ = \frac{\pi}{4}\)
\(60^\circ = \frac{\pi}{3}\)
\(90^\circ = \frac{\pi}{2}\)
\(\sin\theta\)
0
\(\frac{1}{2}\)
\(\frac{\sqrt{2}}{2}\)
\(\frac{\sqrt{3}}{2}\)
1
\(\cos\theta\)
1
\(\frac{\sqrt{3}}{2}\)
\(\frac{\sqrt{2}}{2}\)
\(\frac{1}{2}\)
0
\(\tan\theta\)
0
\(\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\)
1
\(\sqrt{3}\)
undefined
Worked example Combining exact values
Without a calculator, find the exact value of \(2\sin 60^\circ\cos 30^\circ + \tan^2 30^\circ\).
\(\tan^2 30^\circ\) is \((\tan 30^\circ)^2 = \dfrac{1}{3}\), not \(\tan(900^\circ)\) and not \(\tan 30^\circ\) doubled.
Section 02
Related angles of any size.
Every angle has a basic acute angle (its angle to the \(x\)-axis) and a sign that depends on the quadrant. The ratio has the size of the acute angle's ratio and the sign of its quadrant: all positive in the first, only sine in the second, only tangent in the third, only cosine in the fourth.
Second quadrant (\(180^\circ - \alpha\), or \(\pi - \alpha\)): \(\sin\) positive; \(\cos\), \(\tan\) negative.
Third quadrant (\(180^\circ + \alpha\), or \(\pi + \alpha\)): \(\tan\) positive; \(\sin\), \(\cos\) negative.
Fourth quadrant (\(360^\circ - \alpha\), or \(2\pi - \alpha\), or \(-\alpha\)): \(\cos\) positive; \(\sin\), \(\tan\) negative.
Worked example Three related angles
Find the exact values of (a) \(\cos 150^\circ\) (b) \(\sin\dfrac{4\pi}{3}\) (c) \(\tan\dfrac{7\pi}{4}\).
Find the exact values of \(\sin\left(-\dfrac{\pi}{6}\right)\) and \(\cos\left(-\dfrac{5\pi}{3}\right)\).
\(-\dfrac{\pi}{6}\) is measured clockwise into the fourth quadrant, where sine is negative: \(\sin\left(-\dfrac{\pi}{6}\right) = -\tfrac{1}{2}\).
Cosine is symmetric, \(\cos(-x) = \cos x\), so \(\cos\left(-\dfrac{5\pi}{3}\right) = \cos\dfrac{5\pi}{3}\).
\(\dfrac{5\pi}{3} = 2\pi - \dfrac{\pi}{3}\) is in the fourth quadrant, where cosine is positive: \(\cos\dfrac{\pi}{3} = \tfrac{1}{2}\).
Answer: \(-\tfrac{1}{2}\) and \(\tfrac{1}{2}\)
Your turn
Find the exact value of \(\tan\dfrac{5\pi}{4} - 2\cos\dfrac{7\pi}{6}\).
\(\tan\dfrac{5\pi}{4} = \tan\dfrac{\pi}{4} = 1\) (third quadrant, tan positive); \(\cos\dfrac{7\pi}{6} = -\dfrac{\sqrt{3}}{2}\) (third quadrant, cos negative). So \(1 - 2\left(-\dfrac{\sqrt{3}}{2}\right) = 1 + \sqrt{3}\).
Examiner trap
Find the acute angle to the \(x\)-axis, not to the \(y\)-axis. \(150^\circ\) is \(30^\circ\) from the axis (\(180^\circ - 150^\circ\)), not \(60^\circ\) (\(150^\circ - 90^\circ\)).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
What is the exact value of \(\cos 60^\circ\)?
Quick check+10 XP
What is the exact value of \(\tan\dfrac{\pi}{3}\)?
Quick check+10 XP
What is the exact value of \(\cos 150^\circ\)?
Quick check+10 XP
Which of these is an identity, true for every \(\theta\)?
Section 03
The two identities.
An identity (symbol \(\equiv\)) is true for every value of \(\theta\). These two come straight from the unit circle: a point at angle \(\theta\) is \((\cos\theta, \sin\theta)\), its gradient from the origin is \(\dfrac{\sin\theta}{\cos\theta}\), and Pythagoras gives \(\cos^2\theta + \sin^2\theta = 1\).
\(\sin^2\theta + \cos^2\theta \equiv 1\), also used as \(\sin^2\theta \equiv 1 - \cos^2\theta\) and \(\cos^2\theta \equiv 1 - \sin^2\theta\).
Proving an identity: start from one side (usually the more complicated) and transform it into the other. Never treat it as an equation and do the same thing to both sides.
Worked example One ratio gives the others
Given that \(\sin\theta = \dfrac{5}{13}\) and \(\theta\) is obtuse, find the exact values of \(\cos\theta\) and \(\tan\theta\).
After \(\cos^2\theta = \dfrac{144}{169}\), the square root has two signs. The quadrant decides which one; dropping the minus sign for an obtuse angle loses the marks.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
Given that \(\sin\theta = \tfrac{2}{3}\) and \(\theta\) is obtuse, find the exact value of \(\tan\theta\).
Exam style+20 XP
Find the exact value of \(\cos\dfrac{5\pi}{6} - 2\sin\dfrac{5\pi}{3}\).
Exam style+20 XP
Simplify \(\sin\theta\tan\theta + \cos\theta\).
Exam style+20 XP
Given that \(\cos\theta = k\), where \(\dfrac{3\pi}{2} < \theta < 2\pi\), express \(\tan\theta\) in terms of \(k\).
Skill unlocked
Exact values and identities, mastered.
This skill is now recorded on your profile. Keep the chain going.