Types of Trigonometric Identities: List, Proofs and Examples
Types of Trigonometric Identities: Complete List with Proofs and Examples
Trigonometric identities are equations involving trigonometric functions that are true for every value of the angle for which both sides are defined. For example, sin²θ + cos²θ = 1 holds for any angle θ. Identities are the main tool for simplifying expressions, proving results, solving trigonometric equations and evaluating integrals in calculus.
This guide covers all 13 types of trigonometric identities students need for FSc, BS, MSc, PPSC, FPSC, entry tests and other exams, with formula boxes, proofs, solved examples, a values table and practice MCQs with explanations.
The 13 types of trigonometric identities at a glance
- Reciprocal identities
- Quotient (ratio) identities
- Pythagorean identities
- Even–odd (negative angle) identities
- Co-function (complementary angle) identities
- Supplementary, allied and periodic identities
- Sum and difference identities
- Double angle identities
- Triple angle identities
- Half angle identities
- Power-reducing identities
- Product-to-sum identities
- Sum-to-product identities
Also included: triangle identities (laws of sines and cosines), table of values, how to prove identities, solved examples, MCQs and FAQs.
1. Reciprocal Identities
In a right triangle, sin θ = perpendicular/hypotenuse, cos θ = base/hypotenuse and tan θ = perpendicular/base. The other three ratios are their reciprocals.
sec θ = 1 / cos θ | cos θ = 1 / sec θ
cot θ = 1 / tan θ | tan θ = 1 / cot θ
Use: rewrite csc, sec and cot in terms of sin and cos before simplifying. This is the first step in most proofs.
2. Quotient (Ratio) Identities
cot θ = cos θ / sin θ (sin θ ≠ 0)
Why it works: sin θ / cos θ = (perpendicular/hypotenuse) ÷ (base/hypotenuse) = perpendicular/base = tan θ.
3. Pythagorean Identities
1 + tan²θ = sec²θ
1 + cot²θ = csc²θ
Proof: In a right triangle with perpendicular p, base b and hypotenuse h, Pythagoras' theorem gives p² + b² = h².
- Divide by h²: (p/h)² + (b/h)² = 1, so sin²θ + cos²θ = 1.
- Divide by b²: (p/b)² + 1 = (h/b)², so tan²θ + 1 = sec²θ.
- Divide by p²: 1 + (b/p)² = (h/p)², so 1 + cot²θ = csc²θ.
Useful rearranged forms: sin²θ = 1 − cos²θ, cos²θ = 1 − sin²θ, sec²θ − tan²θ = 1, csc²θ − cot²θ = 1.
4. Even–Odd (Negative Angle) Identities
cos(−θ) = cos θ | sec(−θ) = sec θ
tan(−θ) = −tan θ | cot(−θ) = −cot θ
Cosine and secant are even functions (their graphs are symmetric about the y-axis). The other four are odd functions.
5. Co-function (Complementary Angle) Identities
Two angles are complementary when they add up to 90° (π/2).
tan(90° − θ) = cot θ | cot(90° − θ) = tan θ
sec(90° − θ) = csc θ | csc(90° − θ) = sec θ
Example: sin 60° = cos 30° = √3/2, and tan 20° · tan 70° = tan 20° · cot 20° = 1.
6. Supplementary, Allied and Periodic Identities
Two angles are supplementary when they add up to 180° (π). Allied angles are angles such as 90° ± θ, 180° ± θ, 270° ± θ and 360° ± θ.
| Angle | sin | cos | tan |
|---|---|---|---|
| 90° + θ | cos θ | −sin θ | −cot θ |
| 180° − θ | sin θ | −cos θ | −tan θ |
| 180° + θ | −sin θ | −cos θ | tan θ |
| 270° − θ | −cos θ | −sin θ | cot θ |
| 270° + θ | −cos θ | sin θ | −cot θ |
| 360° − θ | −sin θ | cos θ | −tan θ |
Shortcut rule: with 90° or 270°, the function changes (sin ↔ cos, tan ↔ cot); with 180° or 360°, it stays the same. The sign is the sign of the original function in the quadrant where the angle lies (remember "All Students Take Calculus": all positive in QI, sin in QII, tan in QIII, cos in QIV).
Need to switch between degrees and radians (e.g. 180° = π)? Use the free PakMath unit converter.
7. Sum and Difference Identities
sin(A − B) = sin A cos B − cos A sin B
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
tan(A + B) = (tan A + tan B) / (1 − tan A tan B)
tan(A − B) = (tan A − tan B) / (1 + tan A tan B)
Memory tip: the sine formulas keep the sign of the angle; the cosine formulas flip it.
Example: sin 75° = sin(45° + 30°) = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.
8. Double Angle Identities
Put A = B = θ in the sum identities:
cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ = (1 − tan²θ) / (1 + tan²θ)
tan 2θ = 2 tan θ / (1 − tan²θ)
Tip: choose the form of cos 2θ that matches what the question gives you: use 1 − 2sin²θ if you know sin θ, and 2cos²θ − 1 if you know cos θ.
9. Triple Angle Identities
cos 3θ = 4 cos³θ − 3 cos θ
tan 3θ = (3 tan θ − tan³θ) / (1 − 3 tan²θ)
Proof of sin 3θ: sin 3θ = sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ = 2 sin θ cos²θ + (1 − 2sin²θ) sin θ = 2 sin θ(1 − sin²θ) + sin θ − 2sin³θ = 3 sin θ − 4 sin³θ.
10. Half Angle Identities
cos(θ/2) = ±√[(1 + cos θ) / 2]
tan(θ/2) = ±√[(1 − cos θ) / (1 + cos θ)] = sin θ / (1 + cos θ) = (1 − cos θ) / sin θ
Choosing the sign: the ± sign depends on the quadrant of θ/2, not of θ. For example, if θ/2 lies in the second quadrant, sin(θ/2) is positive and cos(θ/2) is negative.
11. Power-Reducing Identities
These come from rearranging the double angle formulas for cos 2θ. They are used heavily in integration (e.g. ∫sin²x dx) and in vector calculus; try our vector analysis calculators to practise related problems.
cos²θ = (1 + cos 2θ) / 2
tan²θ = (1 − cos 2θ) / (1 + cos 2θ)
sin³θ = (3 sin θ − sin 3θ) / 4 | cos³θ = (3 cos θ + cos 3θ) / 4
12. Product-to-Sum Identities
Add or subtract the sum and difference formulas to turn a product into a sum:
2 cos A sin B = sin(A + B) − sin(A − B)
2 cos A cos B = cos(A + B) + cos(A − B)
2 sin A sin B = cos(A − B) − cos(A + B)
13. Sum-to-Product Identities
Put A + B = C and A − B = D in the product-to-sum formulas:
sin C − sin D = 2 cos[(C + D)/2] sin[(C − D)/2]
cos C + cos D = 2 cos[(C + D)/2] cos[(C − D)/2]
cos C − cos D = −2 sin[(C + D)/2] sin[(C − D)/2]
Bonus: Triangle Identities (Laws of Sines and Cosines)
For any triangle ABC with sides a, b, c opposite angles A, B, C and circumradius R:
Law of cosines: a² = b² + c² − 2bc cos A, cos A = (b² + c² − a²) / 2bc
Area: Δ = ½ ab sin C
Angle sum: A + B + C = 180°, so sin(A + B) = sin C and cos(A + B) = −cos C
The law of cosines is closely linked to the dot product of vectors: cos θ = (a · b)/(|a||b|). Practise this in our 50 vector algebra MCQs with answers.
Table of Trigonometric Values for Standard Angles
| θ | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | undefined |
| csc θ | undefined | 2 | √2 | 2/√3 | 1 |
| sec θ | 1 | 2/√3 | √2 | 2 | undefined |
| cot θ | undefined | √3 | 1 | 1/√3 | 0 |
How to Prove Trigonometric Identities (Step by Step)
- Start with the more complicated side and work towards the simpler side. Never move terms across the equals sign as if solving an equation.
- Convert everything to sin and cos using the reciprocal and quotient identities.
- Combine fractions using a common denominator.
- Look for Pythagorean patterns such as 1 − sin²θ, sec²θ − 1 or 1 + cot²θ.
- Factor (difference of squares, common factors) or multiply by the conjugate, e.g. multiply 1/(1 − sin θ) by (1 + sin θ)/(1 + sin θ).
- Expand double or compound angles into single angles when the two sides use different angles.
Solved Examples
Example 1: Prove that (1 − cos²θ)(1 + cot²θ) = 1.
Solution: 1 − cos²θ = sin²θ and 1 + cot²θ = csc²θ. So the left side = sin²θ · csc²θ = sin²θ · (1/sin²θ) = 1 = right side. ✔
Example 2: Prove that sin θ/(1 + cos θ) + (1 + cos θ)/sin θ = 2 csc θ.
Solution: Take the common denominator sin θ(1 + cos θ):
Left side = [sin²θ + (1 + cos θ)²] / [sin θ(1 + cos θ)] = [sin²θ + 1 + 2cos θ + cos²θ] / [sin θ(1 + cos θ)]
= [2 + 2cos θ] / [sin θ(1 + cos θ)] = 2(1 + cos θ) / [sin θ(1 + cos θ)] = 2/sin θ = 2 csc θ. ✔
Example 3: Find the exact value of sin 75°.
Solution: sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.
Example 4: If sin θ = 3/5 and θ lies in the second quadrant, find sin 2θ and cos 2θ.
Solution: cos θ = −√(1 − 9/25) = −4/5 (negative in QII).
sin 2θ = 2 sin θ cos θ = 2(3/5)(−4/5) = −24/25.
cos 2θ = 1 − 2sin²θ = 1 − 18/25 = 7/25.
Example 5: Use a half angle identity to find cos 15°.
Solution: cos 15° = cos(30°/2) = +√[(1 + cos 30°)/2] (positive because 15° is in QI) = √[(2 + √3)/4] = √(2 + √3)/2 = (√6 + √2)/4.
Example 6: Evaluate sin 75° + sin 15° without a calculator.
Solution: By sum-to-product, sin 75° + sin 15° = 2 sin 45° cos 30° = 2(√2/2)(√3/2) = √6/2.
Common Mistakes to Avoid
- Writing sin(A + B) = sin A + sin B. This is false; use the sum identity.
- Confusing sin²θ (which is (sin θ)²) with sin(θ²).
- Writing sin 2θ = 2 sin θ. The correct identity is sin 2θ = 2 sin θ cos θ.
- Choosing the wrong ± sign in half angle formulas. Check the quadrant of θ/2.
- Cancelling terms across a sum, e.g. (sin θ + cos θ)/sin θ ≠ cos θ.
Trigonometric Identities MCQs with Answers
Test yourself. Tap the button under each question to see the answer with explanation.
Q1. sin²θ + cos²θ is equal to:
A) 0
B) 1
C) 2
D) tan²θ
Show Answer with Explanation
Correct answer: B) 1
Explanation: This is the first Pythagorean identity. It comes from dividing p² + b² = h² by h².
Q2. 1 + tan²θ is equal to:
A) csc²θ
B) cot²θ
C) sec²θ
D) sin²θ
Show Answer with Explanation
Correct answer: C) sec²θ
Explanation: Divide sin²θ + cos²θ = 1 by cos²θ: tan²θ + 1 = sec²θ.
Q3. sin(−θ) is equal to:
A) sin θ
B) −sin θ
C) cos θ
D) −cos θ
Show Answer with Explanation
Correct answer: B) −sin θ
Explanation: Sine is an odd function, so sin(−θ) = −sin θ.
Q4. Which of the following is equal to cos 2θ?
A) 2 sin θ cos θ
B) 1 − 2sin²θ
C) 2 cos θ
D) 1 + 2sin²θ
Show Answer with Explanation
Correct answer: B) 1 − 2sin²θ
Explanation: cos 2θ = cos²θ − sin²θ = (1 − sin²θ) − sin²θ = 1 − 2sin²θ.
Q5. sin 3θ is equal to:
A) 3 sin θ − 4 sin³θ
B) 4 sin³θ − 3 sin θ
C) 3 sin θ + 4 sin³θ
D) 3 sin θ
Show Answer with Explanation
Correct answer: A) 3 sin θ − 4 sin³θ
Explanation: sin 3θ = sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ, which simplifies to 3 sin θ − 4 sin³θ.
Q6. tan(A + B) is equal to:
A) (tan A − tan B)/(1 + tan A tan B)
B) (tan A + tan B)/(1 + tan A tan B)
C) (tan A + tan B)/(1 − tan A tan B)
D) tan A + tan B
Show Answer with Explanation
Correct answer: C) (tan A + tan B)/(1 − tan A tan B)
Explanation: Divide sin(A + B) by cos(A + B) and then divide the numerator and denominator by cos A cos B.
Q7. 2 sin A cos B is equal to:
A) sin(A + B) + sin(A − B)
B) sin(A + B) − sin(A − B)
C) cos(A + B) + cos(A − B)
D) cos(A − B) − cos(A + B)
Show Answer with Explanation
Correct answer: A) sin(A + B) + sin(A − B)
Explanation: Adding sin(A + B) = sin A cos B + cos A sin B and sin(A − B) = sin A cos B − cos A sin B gives 2 sin A cos B.
Q8. The value of cos²15° − sin²15° is:
A) 1/2
B) √3/2
C) 1/√2
D) 1
Show Answer with Explanation
Correct answer: B) √3/2
Explanation: cos²θ − sin²θ = cos 2θ, so the value is cos 30° = √3/2.
Q9. sin 150° is equal to:
A) −1/2
B) √3/2
C) 1/2
D) −√3/2
Show Answer with Explanation
Correct answer: C) 1/2
Explanation: sin(180° − θ) = sin θ, so sin 150° = sin 30° = 1/2 (sine is positive in the second quadrant).
Q10. tan 20° · tan 70° is equal to:
A) 0
B) 1
C) √3
D) tan 90°
Show Answer with Explanation
Correct answer: B) 1
Explanation: tan 70° = tan(90° − 20°) = cot 20°, so tan 20° · cot 20° = 1.
Frequently Asked Questions
What are the types of trigonometric identities?
The main types are reciprocal, quotient, Pythagorean, even–odd, co-function, supplementary/allied and periodic, sum and difference, double angle, triple angle, half angle, power-reducing, product-to-sum and sum-to-product identities.
What are the three Pythagorean identities?
sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ and 1 + cot²θ = csc²θ.
What is the difference between a trigonometric identity and a trigonometric equation?
An identity is true for every allowed value of the angle (e.g. sin²θ + cos²θ = 1). An equation is true only for some values (e.g. sin θ = 1/2 is true only for θ = 30°, 150°, … ).
How many trigonometric identities are there?
There are infinitely many, because any identity can be combined into new ones. Students usually need about 40–50 standard formulas, grouped into the 13 types on this page.
Which identities are most important for exams?
The Pythagorean, sum and difference, double angle and half angle identities appear most often, followed by product-to-sum and sum-to-product formulas.
How do I memorise trigonometric identities?
Learn sin²θ + cos²θ = 1 and the sum formulas for sin(A + B) and cos(A + B) well. Almost every other identity (double, triple, half angle, power-reducing and product-to-sum) can be derived from these in one or two steps.
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