Exact Values

Type \sin 41^\circ into a calculator and you get an ugly, endless decimal: 0.6560590\ldots. But a handful of special angles are different. For 0^\circ, 30^\circ, 45^\circ, 60^\circ and 90^\circ the sine, cosine and tangent come out beautifully exact — clean fractions and neat surds like \tfrac12, \tfrac{\sqrt2}{2} and \sqrt3.

These are worth knowing by heart, because exams often forbid a calculator for exactly these angles — and because the whole table comes from just two special triangles, so even if you forget, you can rebuild it in seconds.

The 45-45-90 triangle is a right isosceles triangle: two legs of length 1 and a hypotenuse of \sqrt2.
The 30-60-90 triangle is half an equilateral triangle: sides 1, \sqrt3 and 2.

Reading the side ratios (SOH-CAH-TOA) straight off these triangles:

\sin 45^\circ = \cos 45^\circ = \tfrac{1}{\sqrt2} = \tfrac{\sqrt2}{2}

\sin 30^\circ = \cos 60^\circ = \tfrac12, \qquad \sin 60^\circ = \cos 30^\circ = \tfrac{\sqrt3}{2}

\tan 30^\circ = \tfrac{1}{\sqrt3}, \qquad \tan 45^\circ = 1, \qquad \tan 60^\circ = \sqrt3

The exact trig values at the special angles:

\sin: \sin 0^\circ = 0, \sin 30^\circ = \tfrac12, \sin 45^\circ = \tfrac{\sqrt2}{2}, \sin 60^\circ = \tfrac{\sqrt3}{2}, \sin 90^\circ = 1;
\cos: \cos 0^\circ = 1, \cos 30^\circ = \tfrac{\sqrt3}{2}, \cos 45^\circ = \tfrac{\sqrt2}{2}, \cos 60^\circ = \tfrac12, \cos 90^\circ = 0;
\tan: \tan 0^\circ = 0, \tan 30^\circ = \tfrac{1}{\sqrt3}, \tan 45^\circ = 1, \tan 60^\circ = \sqrt3, \tan 90^\circ is undefined.

A neat memory aid for the sines: write 0, 1, 2, 3, 4 under the angles 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ, then take \tfrac{\sqrt{\phantom{0}}}{2} of each: \tfrac{\sqrt0}{2}, \tfrac{\sqrt1}{2}, \tfrac{\sqrt2}{2}, \tfrac{\sqrt3}{2}, \tfrac{\sqrt4}{2} = 0, \tfrac12, \tfrac{\sqrt2}{2}, \tfrac{\sqrt3}{2}, 1. The cosines are the same list backwards.

Where the values come from

Step through the two triangles. Every exact value above is just one side divided by another, read straight off these pictures.

Worked example 1 — building the 45° values

Take a square of side 1 and cut it along the diagonal. You get a right triangle with two legs of length 1 and — by Pythagoras — a hypotenuse of

\sqrt{1^2 + 1^2} = \sqrt2.

Because both legs are equal, both non-right angles are 45^\circ. Reading off SOH-CAH-TOA:

\sin 45^\circ = \frac{\text{opp}}{\text{hyp}} = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}, \qquad \cos 45^\circ = \frac{\text{adj}}{\text{hyp}} = \frac{1}{\sqrt2} = \frac{\sqrt2}{2}.

And \tan 45^\circ = \tfrac{\text{opp}}{\text{adj}} = \tfrac{1}{1} = 1 — the one everyone remembers.

Worked example 2 — building the 30° and 60° values

Start with an equilateral triangle of side 2 (every angle 60^\circ). Drop a line straight down the middle: it splits the base into two halves of length 1 and cuts the top angle into two 30^\circ halves. Pythagoras gives the height:

\sqrt{2^2 - 1^2} = \sqrt{3}.

Now look at the 30^\circ angle at the top. Its opposite side is the half-base 1, and the hypotenuse is 2:

\sin 30^\circ = \frac{1}{2}, \qquad \cos 30^\circ = \frac{\sqrt3}{2}.

Look instead at the 60^\circ angle at the base: now the tall \sqrt3 side is opposite it, so \sin 60^\circ = \tfrac{\sqrt3}{2} and \cos 60^\circ = \tfrac12. The 30^\circ and 60^\circ values are the same two numbers, just swapped.

Worked example 3 — using them with no calculator

A ladder 10\,\text{m} long leans at exactly 60^\circ to the ground. How high up the wall does it reach — exactly?

The height is opposite the 60^\circ, and the ladder is the hypotenuse, so \text{height} = 10 \times \sin 60^\circ. Because we know the exact value:

\text{height} = 10 \times \frac{\sqrt3}{2} = 5\sqrt3 \approx 8.66\,\text{m}.

The exact answer is 5\sqrt3\,\text{m} — no calculator needed to get there, and no rounding lost along the way.

Keep exact values as surds. If a question asks for the exact value, write \sin 60^\circ = \tfrac{\sqrt3}{2}, not the rounded 0.866. A decimal is only an approximation, and in an "exact value" question it loses marks — the surd is the answer.
Don't muddle the pairs. \sin 30^\circ = \tfrac12 but \sin 60^\circ = \tfrac{\sqrt3}{2} — these are not the same. When in doubt, sketch the half-equilateral triangle: the small angle (30^\circ) faces the short side 1, the big angle (60^\circ) faces the long side \sqrt3. The two special triangles are the reliable way to reconstruct any value you've forgotten.

It really is that economical. Half a square hands you every 45^\circ value; half an equilateral triangle hands you every 30^\circ and 60^\circ value. A couple of Pythagoras sums regenerate the entire table — so there is genuinely nothing to memorise if you can draw those two pictures.

And these particular angles are everywhere in the real world: a 45^\circ ramp, a 60^\circ roof pitch, the 30^\circ–60^\circ set-square in every geometry kit, the hexagons in a honeycomb. That is exactly why they are worth knowing cold. They are also the exact points that anchor the wavy sine and cosine curves — the neat pegs the whole graph is hung on.