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By rohit.pandey1
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Updated on 27 May 2026, 11:40 IST
Trigonometry is one of the most important chapters in CBSE Class 11 Maths. It is not only useful for school exams but also helps in competitive exams like JEE Main and JEE Advanced. Many topics in Class 12 Maths and Physics are also based on trigonometry, so having a strong understanding of formulas is very important.
In this chapter, students learn about trigonometric ratios, identities, angle formulas, and their applications in solving problems. If you know the formulas properly, you can solve questions faster and avoid calculation mistakes in exams.
In this article, we have covered all important Class 11 Trigonometry formulas in a simple and easy-to-understand format. You will find basic trigonometric identities, sum and difference formulas, double angle formulas, triple angle formulas, product-to-sum identities, and quick revision tips — all in one place for easy preparation and revision.
Trigonometry is the branch of mathematics that studies the relationship between angles and sides of triangles using six functions: sine, cosine, tangent, cosecant, secant, and cotangent.
In Class 11, the NCERT chapter Trigonometric Functions (Chapter 3) extends these ratios beyond acute angles to all real numbers, covering:
Why it matters for JEE Main 2026:
Trigonometry doesn't stay confined to its own chapter. You'll use it directly in Calculus (limits, derivatives, integrals), Coordinate Geometry (slopes, angles between lines), Vectors & 3D Geometry, and Physics (SHM, waves, projectile motion, AC circuits). A weak grip on these formulas creates compounding errors across the paper.
Trigonometry typically contributes 2–3 questions in JEE Main Mathematics (both January and April sessions).
| Topic | Expected Questions | Difficulty |
| Trigonometric identities & simplification | 1 | Easy–Medium |
| Inverse trigonometric functions | 1 | Medium |
| Trigonometric equations (general solutions) | 0–1 | Medium–Hard |
Highest-priority topics for JEE Main 2026:
Note:JEE Main 2026 (January session) has already featured questions on compound angle identities and inverse trig domain/range. April session typically repeats similar patterns.
For a right-angled triangle with angle θ, hypotenuse H, opposite side P, and adjacent side B:
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CBSE
| Function | Ratio | Reciprocal |
| sin θ | P / H | cosec θ = 1/sin θ |
| cos θ | B / H | sec θ = 1/cos θ |
| tan θ | P / B | cot θ = 1/tan θ |
Quotient identities:
tan θ = sin θ / cos θcot θ = cos θ / sin θ
Product identities:
sin θ · cosec θ = 1cos θ · sec θ = 1tan θ · cot θ = 1
JEE Tip: These seem elementary, but questions often embed them inside multi-step simplification problems. Practise applying them in reverse — spotting that a complex expression reduces to 1 or 0.
| Angle | 0° | 30° | 45° | 60° | 90° | 180° | 270° | 360° |
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 | 0 | −1 | 0 |
| cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 | −1 | 0 | 1 |
| tan θ | 0 | 1/√3 | 1 | √3 | ∞ | 0 | ∞ | 0 |
| cosec θ | ∞ | 2 | √2 | 2/√3 | 1 | ∞ | −1 | ∞ |
| sec θ | 1 | 2/√3 | √2 | 2 | ∞ | −1 | ∞ | 1 |
| cot θ | ∞ | √3 | 1 | 1/√3 | 0 | ∞ | 0 | ∞ |
sin 0° = √0/2 = 0sin 30° = √1/2 = 1/2sin 45° = √2/2 = 1/√2sin 60° = √3/2sin 90° = √4/2 = 1
Cosine values run in reverse order of sine values across 0°–90°.
The ASTC rule (remembered as "All Students Take Calculus" or "Add Sugar To Coffee") tells you which functions are positive in each quadrant:
| Quadrant | Angle Range | Positive Functions |
| I | 0° – 90° | All (sin, cos, tan, and their reciprocals) |
| II | 90° – 180° | sin θ, cosec θ |
| III | 180° – 270° | tan θ, cot θ |
| IV | 270° – 360° | cos θ, sec θ |
JEE Application: Every time you evaluate an inverse trig function or solve a general equation, you need the correct quadrant sign. This is one of the most common sources of errors in objective questions.
These three identities come from the Pythagorean theorem applied to a unit circle:
sin²θ + cos²θ = 1 ...(i)1 + tan²θ = sec²θ ...(ii) [divide (i) by cos²θ]1 + cot²θ = cosec²θ ...(iii) [divide (i) by sin²θ]
Derived forms (frequently used in JEE):
sin²θ = 1 − cos²θcos²θ = 1 − sin²θsec²θ − tan²θ = 1cosec²θ − cot²θ = 1(sec θ + tan θ)(sec θ − tan θ) = 1(cosec θ + cot θ)(cosec θ − cot θ) = 1
Exam shortcut: When you see (sec θ + tan θ) = k given in a problem, immediately write (sec θ − tan θ) = 1/k. This saves several steps.
Allied angles are angles of the form (nπ/2 ± θ). The rule:
| Expression | Equals |
| sin(90° − θ) | cos θ |
| cos(90° − θ) | sin θ |
| tan(90° − θ) | cot θ |
| sin(90° + θ) | cos θ |
| cos(90° + θ) | −sin θ |
| tan(90° + θ) | −cot θ |
| sin(180° − θ) | sin θ |
| cos(180° − θ) | −cos θ |
| tan(180° − θ) | −tan θ |
| sin(180° + θ) | −sin θ |
| cos(180° + θ) | −cos θ |
| tan(180° + θ) | tan θ |
| sin(270° − θ) | −cos θ |
| cos(270° − θ) | −sin θ |
| sin(270° + θ) | −cos θ |
| cos(270° + θ) | sin θ |
| sin(360° − θ) | −sin θ |
| cos(360° − θ) | cos θ |
sin(A + B) = sin A cos B + cos A sin Bsin(A − B) = sin A cos B − cos A sin Bcos(A + B) = cos A cos B − sin A sin Bcos(A − B) = cos A cos B + sin A sin Btan(A + B) = (tan A + tan B) / (1 − tan A tan B)tan(A − B) = (tan A − tan B) / (1 + tan A tan B)cot(A + B) = (cot A cot B − 1) / (cot B + cot A)cot(A − B) = (cot A cot B + 1) / (cot B − cot A)
Q: Find the value of sin 75°.
sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (1/√2)(√3/2) + (1/√2)(1/2) = √3/(2√2) + 1/(2√2) = (√3 + 1) / (2√2) = (√6 + √2) / 4
2 sin A cos B = sin(A + B) + sin(A − B)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)
sin C + sin D = 2 sin[(C+D)/2] cos[(C−D)/2]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]
JEE Application: Sum-to-product formulas are heavily used in proving identities and in trigonometric equation problems. The cos C − cos D formula is the most commonly missed — note the negative sign.
sin 2A = 2 sin A cos A = 2 tan A / (1 + tan²A)cos 2A = cos²A − sin²A = 1 − 2 sin²A = 2 cos²A − 1 = (1 − tan²A) / (1 + tan²A)tan 2A = 2 tan A / (1 − tan²A)
Derived forms (extremely useful for JEE):
sin²A = (1 − cos 2A) / 2cos²A = (1 + cos 2A) / 2sin A cos A = (sin 2A) / 2
Memory tip: Cosine has three forms for cos 2A. In JEE problems, pick the form that eliminates one function — if the expression has only sin, use cos 2A = 1 − 2sin²A.
sin 3A = 3 sin A − 4 sin³Acos 3A = 4 cos³A − 3 cos Atan 3A = (3 tan A − tan³A) / (1 − 3 tan²A)
Derivation approach: Apply sin(2A + A) or cos(2A + A) and substitute double angle formulas. Knowing the derivation helps you reconstruct the formula under exam pressure.
sin(A/2) = ± √[(1 − cos A) / 2]cos(A/2) = ± √[(1 + cos A) / 2]tan(A/2) = ± √[(1 − cos A) / (1 + cos A)] = sin A / (1 + cos A) = (1 − cos A) / sin A
The ± sign is determined by the quadrant of A/2, not A.
t-substitution (important for JEE integration too):
Let t = tan(A/2). Then:
sin A = 2t / (1 + t²)cos A = (1 − t²) / (1 + t²)tan A = 2t / (1 − t²)
| Function | Domain | Principal Value Range |
| sin⁻¹ x | [−1, 1] | [−π/2, π/2] |
| cos⁻¹ x | [−1, 1] | [0, π] |
| tan⁻¹ x | ℝ | (−π/2, π/2) |
| cosec⁻¹ x | (−∞,−1] ∪ [1,∞) | [−π/2, π/2] − {0} |
| sec⁻¹ x | (−∞,−1] ∪ [1,∞) | [0, π] − {π/2} |
| cot⁻¹ x | ℝ | (0, π) |
Complementary pairs:
sin⁻¹ x + cos⁻¹ x = π/2, |x| ≤ 1tan⁻¹ x + cot⁻¹ x = π/2, x ∈ ℝsec⁻¹ x + cosec⁻¹ x = π/2, |x| ≥ 1
Negative argument:
sin⁻¹(−x) = −sin⁻¹ xcos⁻¹(−x) = π − cos⁻¹ xtan⁻¹(−x) = −tan⁻¹ x
Addition formulas:
tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x+y)/(1−xy)], if xy < 1 = π + tan⁻¹[(x+y)/(1−xy)], if xy > 1, x > 0
Conversion identities:
sin⁻¹ x = cos⁻¹(√(1−x²)) = tan⁻¹(x/√(1−x²))
| Equation | General Solution |
| sin x = 0 | x = nπ |
| cos x = 0 | x = (2n+1)π/2 |
| tan x = 0 | x = nπ |
| sin x = sin α | x = nπ + (−1)ⁿ α |
| cos x = cos α | x = 2nπ ± α |
| tan x = tan α | x = nπ + α |
where n ∈ ℤ (any integer).
Q: Solve: 2 sin²x + sin x − 1 = 0
Factorise: (2 sin x − 1)(sin x + 1) = 0⟹ sin x = 1/2 or sin x = −1Case 1: sin x = 1/2 = sin(π/6) x = nπ + (−1)ⁿ (π/6)Case 2: sin x = −1 = sin(−π/2) x = nπ + (−1)ⁿ (−π/2)
| Mistake | What to Do Instead |
| Writing cos(A+B) = cosA + cosB | Always expand using the full formula |
| Forgetting the ± in half-angle formulas | Check the quadrant of A/2 first |
| Using wrong cos 2A form | Match the form to the existing functions in the expression |
| Missing the condition xy < 1 in tan⁻¹ addition | Write both cases (with and without π) |
| Confusing cos⁻¹(−x) = −cos⁻¹x | It's π − cos⁻¹x, not negative |
| Assuming sin⁻¹(sin x) = x always | True only when x ∈ [−π/2, π/2] |
Based on JEE Main 2022–2026 papers:
| Year/Session | Topic Tested |
| JEE Main Jan 2026 | Compound angle identity, cos 2A simplification |
| JEE Main Apr 2025 | Inverse trig domain + tan⁻¹ addition formula |
| JEE Main Jan 2025 | sin x + sin²x = 1 type — polynomial in cos x |
| JEE Main 2024 | Maximum value of a·sinx + b·cosx expression |
| JEE Main 2023 | General solution with factorisation method |
Pattern observations:
Understand, don't just memorise: Derive sin(A+B) from scratch at least three times. Once you've done that, you can reconstruct it in the exam even if you forget.
Group formulas, don't list them: Memorise compound angle formulas as a group — once you know sin(A+B), you can get sin(A−B) by flipping the middle sign, and cos formulas by shifting phase.
Use active recall: Cover the right side of a formula sheet and try to write the formula from memory. Check. Repeat the ones you missed.
Daily 10-minute formula drill: Write the 6 most-missed formulas from the previous day, from memory, every morning.
Anchor to solved examples: Don't just memorise tan 2A — solve one problem using it immediately. The formula sticks better when attached to a method.
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The NCERT Class 11 syllabus covers approximately 60–70 core formulas across basic ratios, Pythagorean identities, allied angles, compound angles, double/triple/half angles, and inverse trig. You don't need to memorise all of them independently — most are derivable from about 15 base formulas.
Trigonometry (including inverse trig) typically contributes 2–3 questions in JEE Main Maths, worth 8–12 marks. Indirectly, it supports several more questions in calculus and coordinate geometry.
The sum and difference formulas (sin(A±B) and cos(A±B)) are foundational — most other formulas derive from them. Master these first.
Introductory inverse trig (domain, range, principal values) is in Class 11. Detailed properties, graphs, and advanced problems are covered more thoroughly in Class 12. JEE Main tests both.
Express the expression as R·sin(x + φ) or R·cos(x + φ) where R = √(a² + b²). The maximum value is R and the minimum is −R (unless domain restrictions apply).
Focus on the three core forms: sin x = sin α → x = nπ + (−1)ⁿα; cos x = cos α → x = 2nπ ± α; tan x = tan α → x = nπ + α. Almost all trigonometric equations reduce to one of these.