Introduction to Trigonometry

Trigonometry

The branch of mathematics that studies how a triangle's side lengths relate to its angles. Its name comes from Greek words meaning three, sides and measure.

Trigonometric ratios sin A, cos A and tan A

In a right triangle with acute angle A: sin A = opposite side ÷ hypotenuse, cos A = adjacent side ÷ hypotenuse, and tan A = opposite side ÷ adjacent side. 'Opposite' and 'adjacent' are always judged from angle A.

cosec A, sec A and cot A

The reciprocals of the three basic ratios: cosec A = 1/sin A = hypotenuse ÷ opposite, sec A = 1/cos A = hypotenuse ÷ adjacent, and cot A = 1/tan A = adjacent ÷ opposite.

Relation between tan, sin and cos

tan A = sin A / cos A and cot A = cos A / sin A. Both follow from dividing the side ratios for sin A and cos A, since the hypotenuse cancels.

Why do the trigonometric ratios of an angle not change when the right triangle is made bigger or smaller?

Right triangles that share the same acute angle are similar by the AA criterion, so their corresponding sides are proportional. The ratio of any two sides is therefore the same in all of them.

Why can sin A and cos A never be greater than 1, while sec A and cosec A are never less than 1?

The hypotenuse is the longest side of a right triangle, and it is the denominator of sin A and cos A, so they are at most 1. sec A and cosec A are their reciprocals, so they are at least 1.

Given tan A = 4/3, how do you find the other trigonometric ratios of A?

Take the opposite side as 4k and the adjacent side as 3k. Pythagoras theorem gives the hypotenuse as 5k. Then sin A = 4/5 and cos A = 3/5, and the reciprocals give cot A = 3/4, cosec A = 5/4 and sec A = 5/3.

Trigonometric ratios of 45°

sin 45° = cos 45° = 1/√2 and tan 45° = 1. Also cosec 45° = sec 45° = √2 and cot 45° = 1.

Trigonometric ratios of 30° and 60°

sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3. sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3.

Trigonometric ratios of 0° and 90°

sin 0° = 0, cos 0° = 1, tan 0° = 0; sin 90° = 1, cos 90° = 0. cot 0° and cosec 0° are not defined, and neither are tan 90° and sec 90°, because each would mean dividing by zero.

How are the trigonometric ratios of 45° worked out from a triangle?

A right triangle with a 45° angle has its other acute angle 45° too, so the two legs are equal; call each a. Pythagoras theorem gives the hypotenuse as a√2, so sin 45° and cos 45° are both a/(a√2) = 1/√2, and tan 45° = a/a = 1.

How are the trigonometric ratios of 30° and 60° worked out from a triangle?

Take an equilateral triangle of side 2a and drop the perpendicular AD to BC, which bisects it, so BD = a. Pythagoras theorem gives AD = a√3. Triangle ABD has 30° at A and 60° at B, so its sides a, a√3 and 2a give every ratio.

As angle A increases from 0° to 90°, what happens to sin A and cos A?

sin A increases from 0 to 1, because the side opposite A grows towards the hypotenuse. cos A decreases from 1 to 0, because the side adjacent to A shrinks towards zero.

In △ABC right-angled at B, AB = 5 cm and ∠C = 30°. How do you find BC and AC?

Choose the ratio that links the known side with the one you want. AB is opposite C and BC is adjacent, so tan 30° = 5/BC gives BC = 5√3 cm. AB and AC are linked by sin 30° = 5/AC, so AC = 10 cm.

If sin (A − B) = 1/2 and cos (A + B) = 1/2, with A + B ≤ 90° and A > B, how do you find A and B?

Use the specific-angle values: sin 30° = 1/2 gives A − B = 30°, and cos 60° = 1/2 gives A + B = 60°. Solving these two equations together gives A = 45° and B = 15°.

Fundamental trigonometric identity

sin² A + cos² A = 1 for 0° ≤ A ≤ 90°. It comes from dividing AB² + BC² = AC² in a triangle right-angled at B by AC².

Identity involving secant and tangent

1 + tan² A = sec² A, true for 0° ≤ A < 90°. It comes from dividing Pythagoras theorem by AB², the side adjacent to A; at 90° tan A and sec A are not defined.

Identity involving cosecant and cotangent

cot² A + 1 = cosec² A, true for 0° < A ≤ 90°. It comes from dividing Pythagoras theorem by BC², the side opposite A; at 0° cot A and cosec A are not defined.

How do you write cos A, tan A and sec A in terms of sin A?

From sin² A + cos² A = 1, cos A = √(1 − sin² A), taking the positive root because A is acute. Then tan A = sin A / √(1 − sin² A) and sec A = 1 / √(1 − sin² A).

How do you prove that sec A (1 − sin A)(sec A + tan A) = 1?

Write every ratio in terms of sin A and cos A: the left side becomes (1/cos A)(1 − sin A)(1 + sin A)/cos A. The numerator is 1 − sin² A = cos² A, so the whole expression is cos² A / cos² A = 1.

How do you prove (sin θ − cos θ + 1)/(sin θ + cos θ − 1) = 1/(sec θ − tan θ) using sec² θ = 1 + tan² θ?

Divide top and bottom by cos θ to get (tan θ − 1 + sec θ)/(tan θ + 1 − sec θ). Multiply both by (tan θ − sec θ) and replace tan² θ − sec² θ with −1. The common factor (tan θ − sec θ − 1) cancels, leaving 1/(sec θ − tan θ).

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