8 min readConcept sheet1-mark MCQ; 2 or 3-mark use of an identity; the proof of sin²A + cos²A = 1

Trigonometric identities

An identity holds for every angle where both sides are defined. Learn three: sin²A + cos²A = 1, 1 + tan²A = sec²A, 1 + cot²A = cosec²A.

Identity 1sin²A + cos²A = 1, for 0° ≤ A ≤ 90°

Identity 21 + tan²A = sec²A, for 0° ≤ A < 90°

Identity 31 + cot²A = cosec²A, for 0° < A ≤ 90°

Notationsin²A means (sin A)², the square of sin A

Proof of sin²A + cos²A = 1

Take ΔABC with ∠B = 90°, so 0° < A < 90°.

1
By Pythagoras theorem, AB² + BC² = AC².
2
Divide every term by AC²: AB²/AC² + BC²/AC² = AC²/AC².
3
So (AB/AC)² + (BC/AC)² = 1.
4★
AB/AC = cos A and BC/AC = sin A, so cos²A + sin²A = 1.
5
At A = 0°: 0 + 1 = 1. At A = 90°: 1 + 0 = 1. So it holds for 0° ≤ A ≤ 90°.

★ key step

Two more identities from the first

Divide sin²A + cos²A = 1 by	Result	True for
cos²A	tan²A + 1 = sec²A	0° ≤ A < 90° (tan 90° and sec 90° are not defined)
sin²A	1 + cot²A = cosec²A	0° < A ≤ 90° (cot 0° and cosec 0° are not defined)

Rearranged forms you will use

sin²A = 1 − cos²A and cos²A = 1 − sin²A.
sec²A − tan²A = 1, so (sec A − tan A)(sec A + tan A) = 1.
cosec²A − cot²A = 1, so (cosec A − cot A)(cosec A + cot A) = 1.
For an acute angle, take the positive square root: cos A = √(1 − sin²A).

Finding cos A and tan A from sin A

Problem

If sin A = 3/5 and A is acute, find cos A and tan A using an identity.

Working

Identity: sin²A + cos²A = 1, so cos²A = 1 − sin²A.
cos²A = 1 − (3/5)² = 1 − 9/25 = 16/25.
A is acute, so cos A is positive: cos A = 4/5.
tan A = sin A/cos A = (3/5) ÷ (4/5) = 3/4.

Answer

cos A = 4/5, tan A = 3/4

Using sec²θ − tan²θ = 1

Problem

If sec θ + tan θ = 3, find sec θ and tan θ.

Working

Identity: sec²θ − tan²θ = 1, so (sec θ + tan θ)(sec θ − tan θ) = 1.
3 × (sec θ − tan θ) = 1, so sec θ − tan θ = 1/3.
Add the two equations: 2 sec θ = 3 + 1/3 = 10/3, so sec θ = 5/3.
Subtract them: 2 tan θ = 3 − 1/3 = 8/3, so tan θ = 4/3.
Check: 1 + tan²θ = 1 + 16/9 = 25/9 = sec²θ.

Answer

sec θ = 5/3, tan θ = 4/3

In real life

A mason checking a corner with a 3-4-5 string is using Pythagoras theorem. sin²A + cos²A = 1 is the same fact written in ratios, so it never fails.

Exam angle

1 marks: “Find the value of 5 cosec²θ − 5 cot²θ.” 5(cosec²θ − cot²θ) = 5 × 1 = 5.

2 marks: “If cos A = 7/25, find sin A and tan A using an identity.” sin²A = 1 − 49/625 = 576/625, so sin A = 24/25. tan A = (24/25) ÷ (7/25) = 24/7.

3 marks: “If cosec θ + cot θ = 5, find cosec θ and cot θ.” cosec²θ − cot²θ = 1 gives cosec θ − cot θ = 1/5. Adding: 2 cosec θ = 26/5, so cosec θ = 13/5. Then cot θ = 5 − 13/5 = 12/5.

Remember it

One identity makes three. Divide sin²A + cos²A = 1 by cos²A for the sec-tan identity. Divide it by sin²A for the cosec-cot identity.