Class 10 Mathematics · Chapter 8 NotesIntroduction to Trigonometry

Take a right triangle ABC, right-angled at B, and look at the acute angle A. The side BC faces angle A, so it is called the side opposite to angle A. The side AB forms one arm of angle A, so it is called the side adjacent to angle A. The side AC, opposite the right angle, is the hypotenuse. Using these three sides we define six ratios: sine, cosine, tangent and their reciprocals cosecant, secant and cotangent. These are written in short as sin A, cos A, tan A, cosec A, sec A and cot A. Notice that the names of the opposite and adjacent sides change if you shift your attention from angle A to angle C, so always fix the angle first.

• sin A = side opposite to angle A ÷ hypotenuse
• cos A = side adjacent to angle A ÷ hypotenuse
• tan A = side opposite to angle A ÷ side adjacent to angle A
• cosec A = 1 ÷ sin A, sec A = 1 ÷ cos A, cot A = 1 ÷ tan A
• tan A = sin A ÷ cos A and cot A = cos A ÷ sin A

A natural question is whether the ratios change when the triangle becomes bigger or smaller. Suppose you take any point P on the hypotenuse AC of a right triangle ABC and drop a perpendicular PM onto AB. The small triangle PAM is similar to the original triangle CAB by the AA similarity criterion, so their corresponding sides are in the same proportion. Working through the ratios shows that sin A, cos A and tan A have exactly the same values in triangle PAM as in triangle CAB. The same holds if the point lies on AC extended. This proves an important fact: for a fixed angle, the trigonometric ratios do not depend on the lengths of the sides of the triangle.

• Similar triangles have proportional corresponding sides
• Ratios of an angle are fixed numbers, not side lengths
• The Greek letter θ (theta) is often used to name an angle

If even one trigonometric ratio of an acute angle is known, the other five can be found. The method is to treat the given ratio as a ratio of two sides. For example, if tan A = 4/3, take the opposite side as 4k and the adjacent side as 3k, where k is any positive number. The hypotenuse is then found using the Pythagoras theorem: hypotenuse² = (4k)² + (3k)² = 25k², so the hypotenuse is 5k. Once all three sides are known in terms of k, write each ratio using its definition; the k cancels out. Since the hypotenuse is the longest side, sin A and cos A are always less than 1, while sec A and cosec A are always greater than or equal to 1.

• Use Pythagoras theoremhypotenuse² = base² + perpendicular²
• Take the sides as multiples of a positive number k
• sin A and cos A never exceed 1

Some angles appear so often that their ratio values are worth remembering. For 45°, take a right triangle with both legs equal to a; the hypotenuse is a√2, giving sin 45° = cos 45° = 1/√2 and tan 45° = 1. For 30° and 60°, start with an equilateral triangle of side 2a and draw a perpendicular from one vertex to the opposite side. This creates a right triangle with sides a, a√3 and 2a, from which sin 30° = 1/2, cos 30° = √3/2, sin 60° = √3/2 and cos 60° = 1/2. The values for 0° and 90° come from imagining the acute angle shrinking towards 0° or growing towards 90°: sin 0° = 0, cos 0° = 1, sin 90° = 1 and cos 90° = 0. As the angle increases from 0° to 90°, sin A increases from 0 to 1 while cos A decreases from 1 to 0.

• sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, sin 90° = 1
• cos 0° = 1, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, cos 90° = 0
• tan 0° = 0, tan 30° = 1/√3, tan 45° = 1, tan 60° = √3, tan 90° is not defined
• cosec 0° and cot 0° are not defined; sec 90° and tan 90° are not defined

The standard values let you find unknown sides and angles. In a right triangle ABC with the right angle at B, if AB = 5 cm and angle C = 30°, then BC is the side adjacent to angle C and AB is the side opposite to it, so tan 30° = AB ÷ BC. Substituting 1/√3 = 5 ÷ BC gives BC = 5√3 cm. The hypotenuse can then be found using sin 30° = AB ÷ AC, which gives AC = 10 cm, or by the Pythagoras theorem. In another type of problem, if PQ = 3 cm and PR = 6 cm in a right triangle with the right angle at Q, then sin R = PQ ÷ PR = 1/2, so angle R = 30° and angle P = 60°. In general, if one side and any other part of a right triangle are known, the remaining sides and angles can be determined.

• Choose the ratio that connects the known side with the side you want
• The two acute angles of a right triangle add up to 90°

In a right triangle, the two acute angles add up to 90°, so they are complementary. If angle A and angle C are the acute angles, then C = 90° − A. Looking at the definitions, the side opposite to A is the side adjacent to C and vice versa, while the hypotenuse is common. This gives a set of useful relations: sin (90° − A) = cos A, cos (90° − A) = sin A, tan (90° − A) = cot A, cot (90° − A) = tan A, sec (90° − A) = cosec A and cosec (90° − A) = sec A. These relations let you convert a ratio of one angle into a ratio of its complement, which often simplifies calculations.

• sin (90° − A) = cos A and cos (90° − A) = sin A
• tan (90° − A) = cot A and cot (90° − A) = tan A
• sec (90° − A) = cosec A and cosec (90° − A) = sec A

An identity is an equation that is true for all values of the variable for which it is defined. In a right triangle ABC with the right angle at B, the Pythagoras theorem gives AB² + BC² = AC². Dividing every term by AC² gives (AB ÷ AC)² + (BC ÷ AC)² = 1, that is, cos² A + sin² A = 1. Dividing the same equation by AB² gives 1 + tan² A = sec² A, and dividing by BC² gives 1 + cot² A = cosec² A. These three identities are the foundation for proving more complicated identities. To prove an identity, start from one side, usually the more complicated one, and simplify it using the identities until it matches the other side.

• sin² A + cos² A = 1, true for 0° ≤ A ≤ 90°
• 1 + tan² A = sec² A, true for 0° ≤ A < 90°
• 1 + cot² A = cosec² A, true for 0° < A ≤ 90°
• Identities also help express one ratio in terms of another