Trigonometry: Ratios, Results and Identities

This lesson introduces trigonometry through the right triangle: it defines the six trigonometric ratios for an acute angle, builds the reciprocal and square identities from them, and lists the ratio values for the special angles.

Naming the sides of a right triangle

Take a right triangle $ABC$ with the right angle at $B$, so $\angle B = 90^\circ$ and $\angle A$ and $\angle C$ are acute. The side opposite the right angle, $AC$, is the hypotenuse. The other two sides are named relative to whichever acute angle we are looking at.

Relative to angle A

• Hypotenuse: $AC$
• Opposite: $BC$ (the side facing $\angle A$)
• Adjacent: $AB$ (the side next to $\angle A$)

Relative to angle C

• Hypotenuse: $AC$
• Opposite: $AB$
• Adjacent: $BC$

The six trigonometric ratios

For the acute angle $A$, the six ratios are

$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC}, \qquad \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC}, \qquad \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB}$

$\operatorname{cosec} A = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{AC}{BC}, \qquad \sec A = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{AC}{AB}, \qquad \cot A = \frac{\text{adjacent}}{\text{opposite}} = \frac{AB}{BC}$

Always write the name of the angle next to the ratio: a ratio means nothing on its own.

Reciprocal relations

Comparing the definitions, $\sin A$ is opposite over hypotenuse while $\operatorname{cosec} A$ is hypotenuse over opposite, so the two are reciprocals. The same holds for cosine with secant and for tangent with cotangent:

$\operatorname{cosec} A = \frac{1}{\sin A}, \qquad \sec A = \frac{1}{\cos A}, \qquad \cot A = \frac{1}{\tan A}$

Also $\tan A = \dfrac{\sin A}{\cos A}$ and $\cot A = \dfrac{\cos A}{\sin A}$. Multiplying each ratio by its reciprocal gives

$\sin A \cdot \operatorname{cosec} A = 1, \qquad \cos A \cdot \sec A = 1, \qquad \tan A \cdot \cot A = 1$

These are worth knowing by heart, as they come up constantly when proving identities.

The three square identities

The three Pythagorean (square) identities are

$\sin^2 A + \cos^2 A = 1$

$\sec^2 A - \tan^2 A = 1$

$\operatorname{cosec}^2 A - \cot^2 A = 1$

Each can be rearranged by moving a term to the other side. From the first,

$\cos^2 A = 1 - \sin^2 A, \qquad \sin^2 A = 1 - \cos^2 A$

From the second,

$\sec^2 A = 1 + \tan^2 A, \qquad \tan^2 A = \sec^2 A - 1$

From the third,

$\operatorname{cosec}^2 A = 1 + \cot^2 A, \qquad \cot^2 A = \operatorname{cosec}^2 A - 1$

Taking square roots lets us write one ratio in terms of another, for example $\sin A = \sqrt{1 - \cos^2 A}$ and $\sec A = \sqrt{1 + \tan^2 A}$.

Ratios of the special angles

The ratios at $0^\circ, 30^\circ, 45^\circ, 60^\circ$ and $90^\circ$ are

$\sin 0^\circ = 0, \quad \sin 30^\circ = \tfrac{1}{2}, \quad \sin 45^\circ = \tfrac{1}{\sqrt{2}}, \quad \sin 60^\circ = \tfrac{\sqrt{3}}{2}, \quad \sin 90^\circ = 1$

$\cos 0^\circ = 1, \quad \cos 30^\circ = \tfrac{\sqrt{3}}{2}, \quad \cos 45^\circ = \tfrac{1}{\sqrt{2}}, \quad \cos 60^\circ = \tfrac{1}{2}, \quad \cos 90^\circ = 0$

$\tan 0^\circ = 0, \quad \tan 30^\circ = \tfrac{1}{\sqrt{3}}, \quad \tan 45^\circ = 1, \quad \tan 60^\circ = \sqrt{3}, \quad \tan 90^\circ = \text{undefined}$

How to memorise the table

Learn just the sine row, which increases from $0$ to $1$ across the angles. The cosine row is the same values in reverse order, since cosine decreases from $1$ to $0$. The tangent row is then $\tan A = \dfrac{\sin A}{\cos A}$, and the remaining three ratios are the reciprocals of these. Where a denominator is $0$ the ratio is undefined (treated as infinity), since $\tfrac{1}{0} \to \infty$ and $\tfrac{1}{\infty} \to 0$.

Key takeaways

• The six ratios are fixed by which sides count as opposite, adjacent and hypotenuse for the chosen acute angle.
• Cosecant, secant and cotangent are the reciprocals of sine, cosine and tangent.
• The square identities $\sin^2 A + \cos^2 A = 1$, $\sec^2 A - \tan^2 A = 1$ and $\operatorname{cosec}^2 A - \cot^2 A = 1$ rearrange to express any ratio in terms of another.
• Sine rises and cosine falls across the special angles; memorise the sine row and derive the rest.