Trigonometric Ratios and Exact Values

Definition of Sine, Cosine, and Tangent

In any right angled triangle, we define the sides relative to a specific internal angle. Consider the triangle $ABC$ below, where the angle at $C$ is $x^{\circ}$.

1. The side $AB$ is the opposite side because it is directly across from the angle $x^{\circ}$.
2. The side $AC$ is the hypotenuse, which is the longest side of the triangle and sits opposite the right angle.
3. The side $BC$ is the adjacent side, as it is next to the angle $x^{\circ}$ and the right angle.

The three primary trigonometric ratios are defined as follows:

$\sin x^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}}$

$\cos x^{\circ} = \frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan x^{\circ} = \frac{\text{opposite}}{\text{adjacent}}$

Exact Values for 30 and 60 Degrees

You must be able to recall or derive the trigonometric ratios for $30^{\circ}$ and $60^{\circ}$. These are derived from an equilateral triangle with side lengths of 2 units. By splitting this triangle in half with a perpendicular line, we create a right angled triangle with a base of 1 unit and a hypotenuse of 2 units. Using Pythagoras' theorem, the height is $\sqrt{2^{2} - 1^{2}} = \sqrt{3}$.

The values are:

$\sin 30^{\circ} = \cos 60^{\circ} = \frac{1}{2}$

$\sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$

$\tan 30^{\circ} = \frac{1}{\sqrt{3}}$

$\tan 60^{\circ} = \sqrt{3}$

Exact Values for 45 Degrees

The ratios for $45^{\circ}$ are derived from an isosceles right angled triangle where the two equal sides are 1 unit long. By Pythagoras' theorem, the hypotenuse is $\sqrt{1^{2} + 1^{2}} = \sqrt{2}$.

The values are:

$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$

$\tan 45^{\circ} = 1$

Exact Values for 0 and 90 Degrees

Candidates must also know the limiting values of these ratios as the angle approaches $0^{\circ}$ or $90^{\circ}$:

$\sin 90^{\circ} = \cos 0^{\circ} = 1$

$\sin 0^{\circ} = \cos 90^{\circ} = 0$

$\tan 0^{\circ} = 0$

Relationship Between Sine and Cosine

In a right angled triangle $ABC$ where the angle at $B$ is $90^{\circ}$ and the angle at $A$ is $x^{\circ}$, we can find an expression for $\cos x^{\circ}$.

$\cos x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC}$

It is important to note that this is identical to calculating the sine of the other non right angle in the triangle. Since the sum of angles is $180^{\circ}$, the angle at $C$ is $90 - x$. Therefore:

$\sin(90 - x) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC}$

This shows that $\cos x = \sin(90 - x)$.

Worked Example: Deriving Ratios for 30 and 60 Degrees

Problem: Show that $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.

Solution: Imagine an equilateral triangle $ABC$ with side lengths of 2. Let $D$ be a point on $BC$ such that $BD = DC = 1$. The line $AD$ is a line of symmetry, meaning the angle $BAD$ is $30^{\circ}$, the angle $BDA$ is $90^{\circ}$, and the angle $ABD$ is $60^{\circ}$.

Apply Pythagoras' theorem to triangle $ABD$ to find the length $AD$: $AD^{2} = AB^{2} - BD^{2} = 2^{2} - 1^{2} = 3$. So, $AD = \sqrt{3}$.

Now, calculate the sine of $60^{\circ}$: $\sin 60^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AD}{AB} = \frac{\sqrt{3}}{2}$.

Worked Example: Deriving Ratios for 45 Degrees

Problem: Show that $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$.

Solution: Let $ABC$ be an isosceles triangle where the angle $ABC = 90^{\circ}$ and $AB = BC = 1$.

Because the triangle is isosceles, the other two angles are equal: $\frac{180 - 90}{2} = 45^{\circ}$. By Pythagoras' theorem, the hypotenuse $AC$ is: $AC^{2} = AB^{2} + BC^{2} = 1 + 1 = 2$. So, $AC = \sqrt{2}$.

Thus, $\sin 45^{\circ} = \frac{BC}{AC} = \frac{1}{\sqrt{2}}$.

Problems in Two Dimensions

Problem: A circle has a centre $O$ and a radius of 6 cm. $PQ$ is the diameter, and $R$ is a point on the circumference. $PR$ is 8 cm. Find the value of $\sin RPQ$.

Solution: First, draw a diagram.

Since $PQ$ is the diameter, the angle $PRQ$ must be $90^{\circ}$ because it is the angle in a semicircle. The diameter $PQ$ is $2 \times 6 = 12$ cm. We need $\sin RPQ = \frac{RQ}{PQ}$.

Using Pythagoras' theorem to find $RQ$: $RQ^{2} = PQ^{2} - PR^{2} = 12^{2} - 8^{2} = 144 - 64 = 80$. $RQ = \sqrt{80} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5}$.

Therefore, $\sin RPQ = \frac{4\sqrt{5}}{12} = \frac{\sqrt{5}}{3}$.

Problems in Three Dimensions

Problem: $PQRST$ is a square based pyramid with a base $PQRS$ of side 10 cm and a vertical height of 8 cm. What is the sine of the angle that the side $PTQ$ makes with the base $PQRS$?

Solution: Draw a diagram and identify the relevant triangle. Let $O$ be the centre of the square base and $M$ be the midpoint of the side $PQ$. The angle we need is the angle between the face $PTQ$ and the base, which is angle $TMO$.

The distance $OM$ is half the side of the square: $OM = 5$ cm. The vertical height $TO$ is 8 cm. To find $\sin TMO$, we need the hypotenuse $TM$ of the triangle $TOM$.

Using Pythagoras' theorem: $TM^{2} = TO^{2} + OM^{2} = 8^{2} + 5^{2} = 64 + 25 = 89$. So, $TM = \sqrt{89}$.

Thus, $\sin TMO = \frac{TO}{TM} = \frac{8}{\sqrt{89}}$.