Exact values for 30°, 45° and 60° · Trigonometric ratios

“Without using tables or a calculator” means you need exact values, usually with surds. You don’t have to memorise a table. Two small triangles give every value.

Exact valuesPick an angle and a ratio

sin 30° = opposite ÷ hypotenuse = 1 ÷ 2 = 1/2

Cut an equilateral triangle of side 2 down the middle. The base halves to 1, the angles are 60° and 30°, and the height is √(2² − 1²) = √3. Every 30° and 60° value comes from the sides 1, √3 and 2.

Key idea

	$30^\circ$	$45^\circ$	$60^\circ$
$\sin$	$\frac12$	$\frac{1}{\sqrt2} = \frac{\sqrt2}{2}$	$\frac{\sqrt3}{2}$
$\cos$	$\frac{\sqrt3}{2}$	$\frac{1}{\sqrt2} = \frac{\sqrt2}{2}$	$\frac12$
$\tan$	$\frac{1}{\sqrt3} = \frac{\sqrt3}{3}$	$1$	$\sqrt3$

If you forget one, sketch the triangle: half a square (sides $1, 1, \sqrt2$ ) or half an equilateral triangle (sides $1, \sqrt3, 2$ ).

Two special trianglesHalf a square and half an equilateral triangle give every exact value

Worked example · WAEC 2015

WAEC 2015 · Paper 2 · Q10 (a)

Without using mathematical tables or calculators, simplify $\dfrac{2\tan60^\circ + \cos30^\circ}{\sin60^\circ}$ .

Write in the exact values
From the half-equilateral triangle:
- ${\tan 60^\circ = \sqrt3}$ ;
- ${\cos 30^\circ = \frac{\sqrt3}{2}}$ ;
- ${\sin 60^\circ = \frac{\sqrt3}{2}}$ .
Think first. What are $\tan 60^\circ$ , $\cos 30^\circ$ and $\sin 60^\circ$ ?
Substitute
$\frac{2\sqrt3 + \frac{\sqrt3}{2}}{\frac{\sqrt3}{2}}$
Simplify
- Multiply the top and bottom by 2 to clear the halves: ${\frac{4\sqrt3 + \sqrt3}{\sqrt3}}$ .
- ${= \frac{5\sqrt3}{\sqrt3} = 5}$ .

Sine of an angle = cosine of its complement

In any right-angled triangle, the two acute angles add up to $90^\circ$ . The side opposite one of them is next to (adjacent to) the other. So:

Complementary anglesThe side opposite θ is adjacent to 90° − θ

This turns a whole family of questions into simple equations. If $\sin A = \cos B$ (with both angles acute), then $A + B = 90^\circ$ .

Worked example · WAEC 2022

WAEC 2022 · Paper 1 · Q29

Given that $\sin(5x - 28)^\circ = \cos(3x - 50)^\circ$ , $0^\circ \le x \le 90^\circ$ , find the value of $x$ .

Use the complement rule
A sine equals a cosine, so the two angles add up to $90^\circ$ .
Think first. If the sine of one angle equals the cosine of another, how are the two angles related?
Form the equation
$(5x - 28) + (3x - 50) = 90$
Solve
- Collect: ${8x - 78 = 90}$ .
- ${8x = 168}$ , so ${x = 21}$ . The answer is C.

Solving a trigonometric equation

To solve something like $5\cos(x + 10^\circ) - 1 = 0$ , treat the whole bracket as one unknown:

rearrange until the ratio is on its own;
use the inverse (or the tables backwards) to find the bracket;
solve for $x$ .

With an exact value you don’t need tables at all.

Your turn

WAEC 2022 · Paper 2 · Q5 (a)✱

(a)
Given that $m = \tan30^\circ$ and $n = \tan45^\circ$ , simplify, without using a calculator, $\dfrac{m - n}{mn}$ , leaving the answer in the form $p + \sqrt q$ .

Worked solution (try it first)

(a)

$m = \tan 30^\circ = \frac{1}{\sqrt3}$ and $n = \tan 45^\circ = 1$ .
So $\frac{m - n}{mn} = \frac{\frac{1}{\sqrt3} - 1}{\frac{1}{\sqrt3}}$ .
Multiply the top and bottom by $\sqrt3$ : $\frac{1 - \sqrt3}{1} = 1 - \sqrt3$ .
This is in the form $p + \sqrt q$ with $p = 1$ and the surd term $-\sqrt3$ .

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