Trigonometric Ratios and their values

In a right-angled triangle, the trigonometric ratios relate the angles to the lengths of the sides. Consider a right triangle ABC, right-angled at B, with angle \(\theta = \angle BAC\). The sides are named as follows:

• Hypotenuse (\(H\)) = side opposite the right angle (AC)
• Base (\(B\)) = side adjacent to angle \(\theta\) (AB)
• Perpendicular (\(P\)) = side opposite to angle \(\theta\) (BC)

The six trigonometric ratios of angle \(\theta\) are defined as:

• \(\sin \theta = \frac{P}{H} = \frac{BC}{AC}\)
• \(\cos \theta = \frac{B}{H} = \frac{AB}{AC}\)
• \(\tan \theta = \frac{P}{B} = \frac{BC}{AB}\)
• \(\cot \theta = \frac{B}{P} = \frac{AB}{BC}\)
• \(\sec \theta = \frac{H}{B} = \frac{AC}{AB}\)
• \(\csc \theta = \frac{H}{P} = \frac{AC}{BC}\)

Note that \(\csc \theta\), \(\sec \theta\), and \(\cot \theta\) are reciprocals of \(\sin \theta\), \(\cos \theta\), and \(\tan \theta\) respectively. Also, \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\).

The values of these ratios depend only on the angle \(\theta\), not on the size of the triangle.

Worked Illustration

Consider the following values of trigonometric ratios for common angles:

Solved Example 1

Given \(\tan A = \frac{4}{3}\), find the other trigonometric ratios of angle \(A\).

Solution:

Let the perpendicular \(P = 4a\) and base \(B = 3a\). Using Pythagoras theorem, hypotenuse \(H\) is:

\[ H = \sqrt{P^2 + B^2} = \sqrt{(4a)^2 + (3a)^2} = \sqrt{16a^2 + 9a^2} = \sqrt{25a^2} = 5a. \]

Now, calculate the ratios:

• \(\sin A = \frac{P}{H} = \frac{4a}{5a} = \frac{4}{5}\)
• \(\cos A = \frac{B}{H} = \frac{3a}{5a} = \frac{3}{5}\)
• \(\cot A = \frac{1}{\tan A} = \frac{3}{4}\)
• \(\sec A = \frac{1}{\cos A} = \frac{5}{3}\)
• \(\csc A = \frac{1}{\sin A} = \frac{5}{4}\)

Practice Set

Level 1 – Easy

• Find \(\sin 30^\circ\), \(\cos 60^\circ\), and \(\tan 45^\circ\).
• In a right triangle, if \(\sin \theta = \frac{3}{5}\), find \(\cos \theta\) and \(\tan \theta\).

Level 2 – Moderate

• Given \(\tan \theta = 2\), find all other trigonometric ratios.
• In a right triangle, the base is 7 cm and the hypotenuse is 25 cm. Find \(\sin \theta\), \(\cos \theta\), and \(\tan \theta\).

Level 3 – Challenging

• Prove that \(\sin \theta + \cos \theta = 1\) is not possible for any acute angle \(\theta\).
• Find the value of \(\theta\) if \(\tan \theta = \frac{5}{12}\) and calculate all other ratios.

Answer Key

Level 1

• \(\sin 30^\circ = \frac{1}{2}\), \(\cos 60^\circ = \frac{1}{2}\), \(\tan 45^\circ = 1\)
• \(\cos \theta = \frac{4}{5}\), \(\tan \theta = \frac{3}{4}\)

Level 2

• \(\sin \theta = \frac{2}{\sqrt{5}}\), \(\cos \theta = \frac{1}{\sqrt{5}}\), \(\cot \theta = \frac{1}{2}\), \(\sec \theta = \sqrt{5}\), \(\csc \theta = \frac{\sqrt{5}}{2}\)
• Perpendicular \(P = \sqrt{25^2 - 7^2} = 24\) cm, \(\sin \theta = \frac{24}{25}\), \(\cos \theta = \frac{7}{25}\), \(\tan \theta = \frac{24}{7}\)

Level 3

• \(\sin \theta + \cos \theta = 1\) is false for acute angles because \(\sin^2 \theta + \cos^2 \theta = 1\) and both \(\sin \theta\) and \(\cos \theta\) are positive and less than 1.
• Using Pythagoras theorem, hypotenuse \(H = 13a\), \(P = 5a\), \(B = 12a\). Ratios: \(\sin \theta = \frac{5}{13}\), \(\cos \theta = \frac{12}{13}\), \(\cot \theta = \frac{12}{5}\), \(\sec \theta = \frac{13}{12}\), \(\csc \theta = \frac{13}{5}\)

Quick Reference

• \(\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\)
• \(\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}\)
• \(\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\)
• \(\cot \theta = \frac{\text{Base}}{\text{Perpendicular}}\)
• \(\sec \theta = \frac{\text{Hypotenuse}}{\text{Base}}\)
• \(\csc \theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}}\)

Glossary

• Hypotenuse: The longest side of a right-angled triangle, opposite the right angle.
• Base: The side adjacent to the angle of interest in a right triangle.
• Perpendicular: The side opposite the angle of interest in a right triangle.
• Acute Angle: An angle less than 90 degrees.
• Trigonometric Ratios: Ratios of sides of a right triangle relative to an angle.

Trigonometric Identities

A trigonometric identity is an equation involving trigonometric ratios that holds true for all values of the variable within its domain.

Consider a right triangle ABC, right-angled at B. By Pythagoras theorem:

\[ AB^2 + BC^2 = AC^2 \]

Dividing both sides by \(AC^2\),

\[ \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1 \]

Or,

\[ \cos^2 A + \sin^2 A = 1 \]

This is the fundamental Pythagorean identity.

Dividing the original equation by \(AB^2\),

\[ 1 + \tan^2 A = \sec^2 A \]

Dividing the original equation by \(BC^2\),

\[ \cot^2 A + 1 = \csc^2 A \]

These identities hold for angles \(0^\circ \leq A \leq 90^\circ\), with domain restrictions where functions are undefined.

Worked Illustration

Express \(\cos A\), \(\tan A\), and \(\sec A\) in terms of \(\sin A\).

Solution:

From the identity,

\[ \cos^2 A = 1 - \sin^2 A \]

\[ \Rightarrow \cos A = \pm \sqrt{1 - \sin^2 A} \]

Taking the positive root for acute angles,

\[ \cos A = \sqrt{1 - \sin^2 A} \]

Then,

\[ \tan A = \frac{\sin A}{\cos A} = \frac{\sin A}{\sqrt{1 - \sin^2 A}} \]

\[ \sec A = \frac{1}{\cos A} = \frac{1}{\sqrt{1 - \sin^2 A}} \]

Solved Example 2

Prove that

\[ \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta} \]

Solution:

Start with LHS:

\[ \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} \]

Rewrite numerator and denominator in terms of \(\tan \theta\) and \(\sec \theta\):

\[ \sin \theta = \tan \theta \cos \theta, \quad \cos \theta = \cos \theta \]

Multiply numerator and denominator by \(\frac{1}{\cos \theta}\):

\[ \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 + \sec \theta} = \frac{\tan \theta + \sec \theta - 1}{\tan \theta + \sec \theta + 1} \]

Multiply numerator and denominator by \(\tan \theta - \sec \theta\):

\[ \frac{(\tan \theta + \sec \theta - 1)(\tan \theta - \sec \theta)}{(\tan \theta + \sec \theta + 1)(\tan \theta - \sec \theta)} \]

Expand numerator:

\[ (\tan^2 \theta - \sec^2 \theta) - (\tan \theta - \sec \theta) \]

Using identity \(\sec^2 \theta - \tan^2 \theta = 1\), numerator becomes:

\[ -1 - \tan \theta + \sec \theta \]

Denominator simplifies to:

\[ (\tan \theta + \sec \theta)^2 - 1^2 = (\tan \theta + \sec \theta)^2 - 1 \]

After simplification, the expression equals:

\[ \frac{-1}{\tan \theta - \sec \theta} = \frac{1}{\sec \theta - \tan \theta} = \text{RHS} \]

Practice Set

Level 1 – Easy

• Verify the identity \(\sin^2 \theta + \cos^2 \theta = 1\) for \(\theta = 30^\circ\).
• Express \(\sec \theta\) in terms of \(\tan \theta\) using the identity \(1 + \tan^2 \theta = \sec^2 \theta\).

Level 2 – Moderate

• Prove that \(1 + \cot^2 \theta = \csc^2 \theta\).
• Express \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).

Level 3 – Challenging

• Prove the identity \(\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}\).
• Derive the identity \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\) using Pythagorean identities.

Answer Key

Level 1

• \(\sin^2 30^\circ + \cos^2 30^\circ = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1\)
• \(\sec \theta = \sqrt{1 + \tan^2 \theta}\)

Level 2

• Using Pythagorean theorem, \(\cot^2 \theta + 1 = \csc^2 \theta\) is true by dividing sides by \(\sin^2 \theta\).
• \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

Level 3

• Proof provided in solved example 2.
• Using \(\cos^2 \theta + \sin^2 \theta = 1\), \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\) follows by definition of double angle.

Quick Reference

• \(\sin^2 \theta + \cos^2 \theta = 1\)
• \(1 + \tan^2 \theta = \sec^2 \theta\)
• \(1 + \cot^2 \theta = \csc^2 \theta\)
• \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
• \(\cot \theta = \frac{\cos \theta}{\sin \theta}\)

Glossary

• Identity: An equation true for all values of the variable within its domain.
• Pythagoras Theorem: In a right triangle, the square of the hypotenuse equals the sum of squares of the other two sides.
• Secant (\(\sec \theta\)): Reciprocal of cosine.
• Cosecant (\(\csc \theta\)): Reciprocal of sine.
• Cotangent (\(\cot \theta\)): Reciprocal of tangent.