The unit circle is one of the most powerful diagrams in mathematics, a single picture that encodes $\sin$, $\cos$, and $\tan$ for every angle that exists, positive, negative, or beyond $360°$. In this lesson you'll review the foundations you'll need for the rest of Module 3: exact values, reference angles, the ASTC rule, and the bridge between degrees and radians.
Without calculating, what is $\sin(150°)$? What about $\cos\!\left(\dfrac{7\pi}{6}\right)$? How does the unit circle help you find these? Write your gut answer before reading on.
There are only two core moves in every exact-value trig question. Lock find the reference angle and apply ASTC for the sign into muscle memory and you can evaluate any standard angle in seconds.
Every standard trig question lives on one of two roads: find the reference angle (the acute angle to the nearest x-axis), then determine the sign from ASTC to decide whether the value is positive or negative.
For any angle $\theta$ measured anticlockwise from the positive x-axis, the point on the unit circle is $(\cos\theta,\, \sin\theta)$. This gives us:
Key exact values to memorise:
| $\theta$ | $0$ | $\dfrac{\pi}{6}$ | $\dfrac{\pi}{4}$ | $\dfrac{\pi}{3}$ | $\dfrac{\pi}{2}$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $1$ |
| $\cos\theta$ | $1$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{1}{2}$ | $0$ |
| $\tan\theta$ | $0$ | $\dfrac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | undef. |
Unit circle definition: = x, = y, = y/x for point () on the unit circle; Exact value table for 0, {6}, {4}, {3}, {2}, or use the 0/2 4/2 trick for
Pause, copy the unit circle definitions into your book.
Quick check: What is the exact value of $\cos\!\left(\dfrac{\pi}{3}\right)$?
We just saw the unit circle gives us coordinates. That raises a question: how do we handle angles beyond 90 where coordinates go negative? This card answers it → the ASTC rule for signs in every quadrant.
For an angle $\theta$ in any quadrant, the reference angle $\alpha$ is the acute angle between the terminal arm and the nearest part of the x-axis:
The ASTC rule (All Stations To Central) tells you the sign:
Only the labelled ratio (and $\tan$ in Q1 too) is positive in each quadrant; all others are negative.
Using reference angles: to find $\sin(5\pi/6)$, identify Q2, reference angle $= \pi - 5\pi/6 = \pi/6$, sine is positive in Q2, so $\sin(5\pi/6) = +\sin(\pi/6) = \frac{1}{2}$.
Reference angles: Q2 = - ; Q3 = - ; Q4 = 2 -; ASTC: All positive in Q1; Sine only in Q2; Tangent only in Q3; Cosine only in Q4
Pause, copy the ASTC rule and reference angles into your book.
Did you get this? True or false: $\sin\!\left(\dfrac{5\pi}{4}\right)$ is negative because $\dfrac{5\pi}{4}$ lies in quadrant 3.
Worked examples · 3 in a row, reveal as you go
Find the exact value of $\sin\!\left(\dfrac{5\pi}{6}\right)$.
Find the exact value of $\tan(210°)$.
Find all angles $\theta \in [0, 2\pi)$ such that $\sin\theta = \dfrac{1}{2}$.
Fill the gap: $\cos\!\left(\dfrac{3\pi}{4}\right) = $ (enter as −√2/2 or −1/√2).
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\cos(300°) = \cos(60°)$ because $300°$ is in Q4 where cosine is positive.
Activities · practice with the ideas
Find the exact value of $\cos\!\left(\dfrac{3\pi}{4}\right)$. Show your reference angle and ASTC reasoning.
Convert $\dfrac{5\pi}{3}$ radians to degrees.
Find all angles $\theta \in [0, 2\pi)$ such that $\sin\theta = \dfrac{1}{2}$.
Find the exact value of $\tan\!\left(\dfrac{4\pi}{3}\right)$.
Explain why $\tan\theta$ has a different sign pattern to both $\sin\theta$ and $\cos\theta$ individually, but is related to both.
Odd one out: Three of these exact values are correct. Which one is NOT?
Earlier you were asked: what is $\sin(150°)$ and $\cos(7\pi/6)$?
$\sin(150°)$: Q2, reference angle $30°$, sine positive in Q2, so $\sin(150°) = +\sin(30°) = \dfrac{1}{2}$.
$\cos\!\left(\dfrac{7\pi}{6}\right)$: $\dfrac{7\pi}{6} = 210°$, Q3, reference angle $\dfrac{\pi}{6}$, cosine is negative in Q3, so $\cos\!\left(\dfrac{7\pi}{6}\right) = -\cos\!\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}$.
And why does $\tan\theta$ have a different sign pattern? Because $\tan = \sin/\cos$, it is positive when both are negative (Q3) or both are positive (Q1), but negative when they have opposite signs (Q2, Q4).
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the exact value of $\cos\!\left(\dfrac{3\pi}{4}\right)$. (2 marks)
Q2. Convert $\dfrac{5\pi}{3}$ radians to degrees. (1 mark)
Q3. Find all angles $\theta \in [0, 2\pi)$ such that $\sin\theta = \dfrac{1}{2}$. (2 marks)
Activity 1: 1. Q2, ref $= \pi/4$, cos negative: $-\sqrt{2}/2$ · 2. $\frac{5\pi}{3} \times \frac{180}{\pi} = 300°$ · 3. Q1 and Q2, ref $= \pi/6$: $\theta = \pi/6, 5\pi/6$ · 4. Q3, ref $= \pi/3$, tan positive: $+\sqrt{3}$ · 5. $\tan = \sin/\cos$; both negative in Q3 makes a positive ratio; opposite signs in Q2 and Q4 make a negative ratio.
Q1 (2 marks): $\frac{3\pi}{4}$ is in Q2; reference angle $= \pi - \frac{3\pi}{4} = \frac{\pi}{4}$ [1]; cos is negative in Q2, so $\cos\!\left(\frac{3\pi}{4}\right) = -\cos\!\left(\frac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}$ [1].
Q2 (1 mark): $\frac{5\pi}{3} \times \frac{180}{\pi} = 300°$ [1].
Q3 (2 marks): $\sin\theta > 0$ in Q1 and Q2 [0.5]; reference angle $= \frac{\pi}{6}$ [0.5]; $\theta = \frac{\pi}{6}$ [0.5] and $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ [0.5].
Odd one out: C is incorrect. $\tan(300°)$: Q4, reference $= 60°$, tan is negative in Q4, so $\tan(300°) = -\sqrt{3}$, not $+\sqrt{3}$.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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