Most angles produce messy decimals for $\sin$, $\cos$ and $\tan$. Five angles — $0°, 30°, 45°, 60°, 90°$ — are exceptions: they come from two simple triangles and give exact values in surd form. Cambridge expects these memorised, so results can be stated exactly, without a calculator, and "show that" questions can be answered algebraically rather than by rounding.

Half a unit square, cut along its diagonal, gives a right-angled triangle with legs $1$ and $1$, and hypotenuse $\sqrt{1^2+1^2}=\sqrt2$:

$\sin 45°$ and $\cos 45°$ are identical because the triangle is isosceles — the opposite and adjacent sides are the same length whichever base angle is chosen.

Half an equilateral triangle of side $2$, split by a vertical line of symmetry, gives sides $1$, $\sqrt3$ and hypotenuse $2$:

$\sin 30°$ matching $\cos 60°$ (and $\sin 60°$ matching $\cos 30°$) isn't a coincidence — the two non-right angles in this triangle add to $90°$, so the "opposite" side for one angle is always the "adjacent" side for the other.

These come from a right-angled triangle collapsing at its limit, not from a triangle you can actually draw. As the angle at a vertex shrinks towards $0°$, the "opposite" side shrinks towards zero length while the hypotenuse and adjacent side become nearly equal — giving $\sin 0°=0$, $\cos 0°=1$, $\tan 0°=0$. Pushing the angle up towards $90°$ swaps those roles: the adjacent side shrinks towards zero instead, giving $\sin 90°=1$, $\cos 90°=0$. Since $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, dividing by the vanishing $\cos 90°=0$ is exactly why $\tan 90°$ is undefined rather than equal to some very large number.

Every tangent value in the table can be rebuilt from the identity:

For example, if $\tan 60°$ is forgotten: $\tan 60° = \dfrac{\sin 60°}{\cos 60°} = \dfrac{\sqrt3/2}{1/2} = \sqrt3$, matching the table exactly. This is a useful safety net that only needs two memorised facts instead of three.

$\tan 90°$ is undefined, not a missing memorisation fact — it corresponds to dividing by $\cos 90° = 0$.

Solve $2\cos x - \sqrt3 = 0$ for $0° \le x \le 90°$, without using a calculator.

Solution

The table shows $\cos 30° = \dfrac{\sqrt3}{2}$, so $x = 30°$.

If the question says "exact value" or "without a calculator," a rounded decimal — even a numerically correct one — will not earn full marks; the surd form itself is what's being tested.

A right-angled triangle has a $60°$ angle and an adjacent side of length $4$ cm. Find the exact length of the opposite side, giving your answer as a surd in its simplest form.

Solution

Because $\tan 60° = \sqrt3$ is exact, the final answer stays as the exact surd $4\sqrt3$ cm rather than being converted to a rounded decimal — always check whether a question is asking for an exact form before reaching for the calculator's decimal display.