The three special angles, from two triangles

Every trigonometry course asks you to know the values of $\sin$ , $\cos$ and $\tan$ at $30^\circ$ , $45^\circ$ and $60^\circ$ by heart. Almost every course also presents them as a table to be memorised, which is why they evaporate two weeks after the exam.

They are not arbitrary numbers. Two figures you already know produce all nine of them, and once you have seen where they come from you can rebuild the table on the back of an exam paper in thirty seconds.

Half an equilateral triangle gives 30° and 60°

Take an equilateral triangle of side $a$ and cut it down its altitude. The altitude of an equilateral triangle is $h = \frac{a\sqrt{3}}{2}$ , so the right triangle you are left with has hypotenuse $a$ , short leg $\frac{a}{2}$ , and long leg $\frac{a\sqrt{3}}{2}$ . The angle at the top is $30^\circ$ ; the one at the base is $60^\circ$ .

Now just read the definitions off that triangle:

\sin 30^\circ = \frac{a/2}{a} = \frac{1}{2} \qquad \cos 30^\circ = \frac{a\sqrt3/2}{a} = \frac{\sqrt3}{2}

\tan 30^\circ = \frac{a/2}{a\sqrt3/2} = \frac{1}{\sqrt3} = \frac{1}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3} = \frac{\sqrt3}{3}

That last step is rationalising the denominator — multiplying above and below by $\sqrt3$ so no root is left underneath. It changes nothing about the value; it is a convention that makes answers comparable.

Read the same triangle from the other acute angle and the opposite and adjacent legs swap roles. That hands you the $60^\circ$ column for free:

\sin 60^\circ = \frac{\sqrt3}{2}, \qquad \cos 60^\circ = \frac12, \qquad \tan 60^\circ = \sqrt3

Half a square gives 45°

A square of side $a$ has diagonal $d = a\sqrt2$ . Cut it corner to corner and the right triangle has two legs of $a$ and a hypotenuse of $a\sqrt2$ , with both acute angles equal to $45^\circ$ . So

\sin 45^\circ = \cos 45^\circ = \frac{a}{a\sqrt2} = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}, \qquad \tan 45^\circ = \frac{a}{a} = 1

The master table

These nine values carry most of a trigonometry course. Learn them once, properly.

	$30^\circ$	$45^\circ$	$60^\circ$
$\sin$	$\dfrac12$	$\dfrac{\sqrt2}{2}$	$\dfrac{\sqrt3}{2}$
$\cos$	$\dfrac{\sqrt3}{2}$	$\dfrac{\sqrt2}{2}$	$\dfrac12$
$\tan$	$\dfrac{\sqrt3}{3}$	$1$	$\sqrt3$
$\cot$	$\sqrt3$	$1$	$\dfrac{\sqrt3}{3}$

The pattern worth noticing. Read the sine row left to right and write each entry over $2$ : $\frac{\sqrt1}{2}, \frac{\sqrt2}{2}, \frac{\sqrt3}{2}$ . The cosine row is the same three numbers backwards.

That symmetry is not a coincidence, and it is not a memory trick either — it is the complementary-angle rule showing up in the table.

Why the rows mirror each other

In a right triangle the two acute angles satisfy $\alpha + \beta = 90^\circ$ . Read the definitions from $\beta$ instead of $\alpha$ and the roles of the legs swap: what was opposite becomes adjacent. So

\sin(90^\circ - \alpha) = \cos\alpha \qquad \cos(90^\circ - \alpha) = \sin\alpha

\tan(90^\circ - \alpha) = \cot\alpha \qquad \cot(90^\circ - \alpha) = \tan\alpha

Since $30^\circ$ and $60^\circ$ are complements of each other, and $45^\circ$ is its own complement, the table has to read the same backwards. This is also where the co- in cosine and cotangent comes from: the complement.

One last thing, because it is the single most common beginner’s error in this subject: $\sin$ on its own means nothing. It is not a quantity you can multiply or cancel — it is the name of a function, and it must always be applied to an angle. There is no cancelling the sines in $\frac{\sin\alpha}{\sin}$ .

special angles
exact values
right triangle

Half an equilateral triangle gives 30° and 60°

Half a square gives 45°

The master table

Why the rows mirror each other

The trigonometry formula sheet