SOHCAHTOA only makes sense for the acute angles inside a right-angled triangle, so it cannot handle an angle of $120^\circ$, a negative angle, or an angle beyond a full turn.
The unit circle fixes this by defining sine and cosine for any angle, using the coordinates of a point moving around a circle of radius 1.
This single idea powers the exact values, the quadrant sign rules, the graphs in SL 3.7, and the equations in SL 3.8.
The Unit Circle Definition
Draw a circle of radius 1 centred at the origin, and let $P$ be the point you reach by turning anticlockwise through angle $\theta$ from the positive $x$-axis.
Then $\cos\theta$ is the $x$-coordinate of $P$ and $\sin\theta$ is the $y$-coordinate of $P$.
Tangent is defined from these: $$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$$ which is also the gradient of the line $OP$.
A line through the origin making angle $\theta$ with the positive $x$-axis therefore has equation $y = x\tan\theta$.
Tip
The signs follow the quadrant of $P$: in the first quadrant all three ratios are positive, in the second only $\sin$, in the third only $\tan$, and in the fourth only $\cos$ (the CAST rule).
Exact Trigonometric Ratios
Two special triangles give every exact value you need: the half-square gives $45^\circ$, and the half-equilateral triangle gives $30^\circ$ and $60^\circ$.
You must be able to recall these instantly for the non-calculator Paper 1.
Notice the sine values run $\tfrac{\sqrt{0}}{2}, \tfrac{\sqrt{1}}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{3}}{2}, \tfrac{\sqrt{4}}{2}$, which is an easy pattern to remember, and cosine is the same list reversed.
Example question
Find the exact value of $\cos\dfrac{3\pi}{4}$.
Solution
The angle $\frac{3\pi}{4}$ is in the second quadrant, where cosine is negative.
Its related acute angle is $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$, and $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$.
Apply the quadrant sign: $$\cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$$
Angles In Different Quadrants
Because the unit circle is symmetric, an angle in any quadrant relates back to an acute angle with only a sign change.
Cosine is even and sine is odd: $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$.
Reflecting across the $y$-axis gives $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$.
Adding half a turn gives $\sin(\pi + x) = -\sin x$ and $\cos(\pi + x) = -\cos x$.
The method for any such value: find the related acute angle, take its exact value, then attach the sign from the quadrant.
The Ambiguous Case Of The Sine Rule
When the sine rule is used to find an angle, remember that $\sin\theta = \sin(180^\circ - \theta)$, so a positive sine gives two possible angles.
This ambiguous case arises with two sides and a non-included angle (SSA), where an obtuse second answer can also form a valid triangle.
Always test the second angle: keep it only if the three angles still add to less than $180^\circ$.
Example question
In triangle $ABC$, $a = 7\,\mathrm{cm}$, $b = 10\,\mathrm{cm}$ and $A = 35^\circ$. Find the two possible values of angle $B$.
Solution
Use the sine rule for an angle: $\frac{\sin B}{b} = \frac{\sin A}{a}$, so $\sin B = \frac{10\sin 35^\circ}{7}$.
Evaluate: $\sin B = 0.819$, giving the acute solution $B = 55.0^\circ$ (3 s.f.).
The obtuse partner is $B = 180^\circ - 55.0^\circ = 125^\circ$.
Check: $35^\circ + 125^\circ = 160^\circ < 180^\circ$, so both triangles are valid.
Active recall
On the unit circle, which coordinate is $\cos\theta$ and which is $\sin\theta$?
State the exact values of $\sin\frac{\pi}{3}$, $\cos\frac{\pi}{4}$ and $\tan\frac{\pi}{6}$.
In which quadrants is each of sine, cosine and tangent positive?
How do you find $\cos\frac{3\pi}{4}$ from a related acute angle?
Why can the sine rule give two possible angles, and how do you decide which are valid?
