In a right-angled triangle an angle must be less than $90°$. But $\sin 200°$ and $\cos 340°$ are perfectly well defined. The trick is to stop thinking about triangles and think instead about a point going round a circle.
with $\theta$ measured anticlockwise from the positive $x$-axis. Then
$\cos\theta$ is the $x$-coordinate of $P$ · $\sin\theta$ is the $y$-coordinate of $P$
$180°$–$270°$: Tangent only · $270°$–$360°$: Cosine only
| Graph | Period | Range | Zeros | Other features |
|---|---|---|---|---|
| $y = \sin x$ | $360°$ | $-1 \leqslant y \leqslant 1$ | $0°, 180°, 360°$ | Max $1$ at $90°$, min $-1$ at $270°$ |
| $y = \cos x$ | $360°$ | $-1 \leqslant y \leqslant 1$ | $90°, 270°$ | Max $1$ at $0°$ and $360°$ |
| $y = \tan x$ | $180°$ | All real values | $0°, 180°, 360°$ | Asymptotes at $90°, 270°$ |
$\sin(-\theta) = -\sin\theta$ · $\cos(-\theta) = \cos\theta$ · $\tan(180° + \theta) = \tan\theta$
When a question asks for an exact answer, a decimal from the calculator will not do. These values come from the two special triangles.
| $\theta$ | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\dfrac12$ | $\dfrac{1}{\sqrt2}$ | $\dfrac{\sqrt3}{2}$ | $1$ |
| $\cos\theta$ | $1$ | $\dfrac{\sqrt3}{2}$ | $\dfrac{1}{\sqrt2}$ | $\dfrac12$ | $0$ |
| $\tan\theta$ | $0$ | $\dfrac{1}{\sqrt3}$ | $1$ | $\sqrt3$ | undefined |
Find the exact value of $\sin 150°$ and of $\cos 150°$.
Find the exact value of $\tan 225°$.
Find all values of $\theta$ in $0° \leqslant \theta \leqslant 360°$ for which $\cos\theta = -\dfrac{1}{2}$.
Definition
$(\cos\theta, \sin\theta)$ is a point on the unit circle.
CAST
All, Sine, Tangent, Cosine — anticlockwise.
Related angle
Find the acute angle, then attach the sign.
$\sin$ symmetry
$\sin(180°-\theta) = \sin\theta$.
$\cos$ symmetry
$\cos(360°-\theta) = \cos\theta$.
Periods
$\sin$, $\cos$: $360°$. $\tan$: $180°$.
Ranges
$\sin$, $\cos$: $[-1,1]$. $\tan$: everything.
$\tan$ asymptotes
Where $\cos\theta = 0$: $90°$, $270°$, …
Exact values
From the $30$–$60$–$90$ and $45$–$45$–$90$ triangles.
Impossible
$\sin\theta = 1.4$ has no solution.
In which quadrants is $\sin\theta$ negative?
▶ Show solution
The third and fourth — that is, $180° < \theta < 360°$.
From CAST, sine is positive only in the first and second quadrants.
Write down the exact value of $\sin 60°$ and $\cos 45°$.
▶ Show solution
$\sin 60° = \dfrac{\sqrt3}{2}$
$\cos 45° = \dfrac{1}{\sqrt2}$ (equivalently $\dfrac{\sqrt2}{2}$)
Find the exact value of $\cos 120°$.
▶ Show solution
Second quadrant, so cosine is negative. Related acute angle $180° - 120° = 60°$.
$\cos 120° = -\cos 60° = -\dfrac{1}{2}$
Find the exact value of $\sin 240°$.
▶ Show solution
Third quadrant, so sine is negative. Related acute angle $240° - 180° = 60°$.
$\sin 240° = -\dfrac{\sqrt3}{2}$
State the period and range of $y = \tan x$.
▶ Show solution
Period $180°$.
Range: all real values — $\tan x$ is unbounded, with asymptotes at $90°$, $270°$ and so on.
Given $\sin 35° = 0.574$, write down the value of $\sin 145°$.
▶ Show solution
$145° = 180° - 35°$, and $\sin(180° - \theta) = \sin\theta$.
$\sin 145° = 0.574$
Find all $\theta$ in $0° \leqslant \theta \leqslant 360°$ with $\sin\theta = -0.5$.
▶ Show solution
Related acute angle: $\sin^{-1}(0.5) = 30°$.
Sine is negative in the third and fourth quadrants:
$180° + 30° = 210°$ and $360° - 30° = 330°$
$\theta = 210°$ or $330°$.
Explain why $\cos\theta = 1.2$ has no solutions.
▶ Show solution
$\cos\theta$ is the $x$-coordinate of a point on the unit circle, so it can never exceed $1$ in magnitude.
The range of $\cos\theta$ is $-1 \leqslant \cos\theta \leqslant 1$, and $1.2$ lies outside it.
If a calculation ever produces this, you have made an arithmetic error earlier.
Sketch $y = \sin x$ for $0° \leqslant x \leqslant 360°$ and use it to explain why $\sin x = 0.8$ has exactly two solutions in that interval.
▶ Show solution
The curve starts at $(0,0)$, rises to a maximum of $1$ at $90°$, falls back through $(180°, 0)$, down to $-1$ at $270°$, and returns to $(360°, 0)$.
The horizontal line $y = 0.8$ lies between $0$ and $1$, so it cuts the rising part of the first hump once and the falling part once — two crossings.
In the second half of the interval the curve is negative, so it never reaches $0.8$ again.
(Solving: $x = 53.1°$ and $x = 180° - 53.1° = 126.9°$.)
$\theta$ is an angle with $\sin\theta = \tfrac{3}{5}$.
(a) If $\theta$ is acute, find the exact values of $\cos\theta$ and $\tan\theta$. (b) If instead $\theta$ is obtuse, find the exact values of $\cos\theta$ and $\tan\theta$. (c) Find both possible values of $\theta$, to 1 d.p. (d) Explain why the question must tell you whether $\theta$ is acute or obtuse.
▶ Show solution
(a) Use $\sin^2\theta + \cos^2\theta = 1$:
$\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$, so $\cos\theta = \pm\dfrac45$.
$\theta$ acute means the first quadrant, where cosine is positive: $\cos\theta = \dfrac{4}{5}$.
$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{4/5} = \dfrac{3}{4}$
(This is the familiar $3$–$4$–$5$ triangle.)
(b) $\theta$ obtuse means the second quadrant, where cosine is negative:
$\cos\theta = -\dfrac{4}{5}$ and $\tan\theta = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}$
Sine is unchanged at $\tfrac35$, consistent with CAST: sine is positive in both quadrants.
(c) $\sin^{-1}(0.6) = 36.9°$ (1 d.p.)
The obtuse solution is $180° - 36.9° = 143.1°$.
So $\theta = 36.9°$ or $143.1°$.
(d) Knowing $\sin\theta$ alone does not determine $\theta$: two angles in $0°$ to $180°$ share the same sine, because of the symmetry $\sin(180° - \theta) = \sin\theta$.
Those two angles have opposite-signed cosines and tangents, so without being told which one is meant, $\cos\theta$ and $\tan\theta$ are genuinely ambiguous.
This is exactly the source of the ambiguous case of the sine rule.