The three trigonometric identities
An identity is true for every allowed angle. Learn the three basic ones: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ.
How to draw the figure
- Draw a right triangle ABC with the right angle at B.
- Mark the angle θ at A.
- Label BC as the opposite side, AB as the adjacent side and AC as the hypotenuse (the longest side, opposite the right angle).
- Write the ratios beside it: sin θ = BC/AC, cos θ = AB/AC, tan θ = BC/AB.
Proving the three identities
An **identity** is an equation that is true for every value of θ for which both sides are defined. Start from **Pythagoras theorem** in the triangle above: AB² + BC² = AC².
- 1★Divide AB² + BC² = AC² by AC²: (AB/AC)² + (BC/AC)² = 1So cos²θ + sin²θ = 1
- 2Divide AB² + BC² = AC² by AB²: 1 + (BC/AB)² = (AC/AB)²So 1 + tan²θ = sec²θ
- 3Divide AB² + BC² = AC² by BC²: (AB/BC)² + 1 = (AC/BC)²So cot²θ + 1 = cosec²θ
- 4Shortcut: divide sin²θ + cos²θ = 1 by cos²θ to get identity 2, or by sin²θ to get identity 3.
★ key step
Other ways to write each identity
| Identity | Rearranged forms |
|---|---|
| sin²θ + cos²θ = 1 | sin²θ = 1 − cos²θ; cos²θ = 1 − sin²θ |
| 1 + tan²θ = sec²θ | sec²θ − tan²θ = 1; (sec θ − tan θ)(sec θ + tan θ) = 1 |
| 1 + cot²θ = cosec²θ | cosec²θ − cot²θ = 1; (cosec θ − cot θ)(cosec θ + cot θ) = 1 |
Example 1: find the other ratios
If sin θ = 3/5 and θ is acute, find cos θ and tan θ.
- Use sin²θ + cos²θ = 1
- cos²θ = 1 − (3/5)² = 1 − 9/25 = 16/25
- cos θ = 4/5 (positive, because θ is acute)
- tan θ = sin θ/cos θ = (3/5) ÷ (4/5) = 3/4
cos θ = 4/5 and tan θ = 3/4
Example 2: how high does the ladder reach?
A 5 m ladder leans against a wall. It makes an angle θ with the ground, and cos θ = 0.6. Find sin θ and the height the ladder reaches on the wall.
- sin²θ = 1 − cos²θ = 1 − 0.36 = 0.64
- sin θ = 0.8 (positive, because θ is acute)
- sin θ = height/length of ladder, so height = 5 × 0.8
- Height = 4 m
sin θ = 0.8, and the ladder reaches 4 m up the wall.
A mason checks a right-angled corner with a string triangle of sides 3 m, 4 m and 5 m. Here sin θ = 3/5 and cos θ = 4/5, and (3/5)² + (4/5)² = 1: the identity at work.
1 marks: “What is the value of cos²θ + 1/(1 + cot²θ)?” 1/(1 + cot²θ) = 1/cosec²θ = sin²θ, so the value is cos²θ + sin²θ = **1**.
2 marks: “If cos θ = 5/13 and θ is acute, find sin θ and tan θ.” sin²θ = 1 − 25/169 = 144/169, so **sin θ = 12/13** and **tan θ = 12/5**.
5 marks: “Prove that sin²θ + cos²θ = 1. Hence prove that 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ.” Draw right triangle ABC with ∠B = 90° and ∠A = θ. From AB² + BC² = AC², divide by AC², then AB², then BC², and write each fraction as a trigonometric ratio.
Start with sin² + cos² = 1. Divide by cos² to get the tan and sec identity; divide by sin² to get the cot and cosec identity.