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Mathematics

Trigonometric Ratios & Identities

📘 Trigonometry · Chapter MN17 🎯 NDA Level : High Priority

Trigonometric Ratios and Identities is one of the highest-yield chapters in NDA Mathematics. The identities — fundamental, compound angle, multiple angle, and product-to-sum — appear directly in simplification questions and as building blocks for every subsequent trigonometry chapter. A student who knows these identities cold can answer 8–10 NDA questions on trig quickly and accurately.

📌 What to expect in NDA (based on 2022–2024 papers):
(1) Conversion between degrees and radians; (2) Finding values of trig ratios at standard angles (0°, 30°, 45°, 60°, 90°); (3) Signs of trig ratios in each quadrant (ASTC rule); (4) Simplifying expressions using fundamental identities (sin²+cos²=1, etc.); (5) Applying sum & difference formulae for sin(A±B), cos(A±B), tan(A±B); (6) Multiple angle formulae: sin 2A, cos 2A in various forms, tan 2A; (7) Sub-multiple angles: sin A/2, cos A/2, tan A/2; (8) Transformation formulae: sum-to-product and product-to-sum conversions.

Topics at a Glance

① Degrees & Radians

Conversion, arc length, sector area

② Trig Ratios & Signs

ASTC rule, quadrant sign chart, standard values

③ Fundamental Identities

sin²+cos²=1, Pythagorean identities

④ Sum & Difference Formulas

sin(A±B), cos(A±B), tan(A±B)

⑤ Multiple & Sub-Multiple Angles

sin 2A, cos 3A, tan A/2, half-angle

⑥ Transformation Formulas

C+D, C−D, product-to-sum

1. Degree & Radian Measure

1.1

Angle Measurement — Degrees, Radians & Conversions

Radians are the natural unit of angle — most calculus and advanced trig uses radians

⚡ Degree ↔ Radian Conversion

Key relationship: π radians = 180° To convert: Degrees → Radians: multiply by π/180 Radians → Degrees: multiply by 180/π Standard conversions (must memorise): 0° = 0 30° = π/6 45° = π/4 60° = π/3 90° = π/2 120° = 2π/3 135° = 3π/4 150° = 5π/6 180° = π 270° = 3π/2 360° = 2π Arc length and sector area (radius r, angle θ in radians): Arc length: l = rθ Sector area: A = (1/2)r²θ = (1/2)rl

Memory trick: 180° = π radians. To go from degrees to radians, multiply by π/180 and simplify. The degree symbol (°) and the π are like conversion labels that cancel out.

2. Trigonometric Ratios — Values & Signs

2.1

Signs in Four Quadrants — ASTC Rule

"All Silly Tom Cats" or "Add Sugar To Coffee" — remembers which ratios are positive

Quadrant II (90°–180°)

sin > 0, cos < 0, tan < 0

Only Sine (and cosec) positive

Quadrant I (0°–90°)

All > 0

All ratios positive

Quadrant III (180°–270°)

tan > 0, sin < 0, cos < 0

Only Tan (and cot) positive

Quadrant IV (270°–360°)

cos > 0, sin < 0, tan < 0

Only Cos (and sec) positive

Reading anti-clockwise from Q4: C → A → S → T — or clockwise from Q1: A → S → T → C.

⚡ Allied Angle Reductions — Quadrant Transformations

Rule: when angle is (n × 90° ± θ): • If n is ODD: sin ↔ cos, tan ↔ cot, sec ↔ cosec (function changes) • If n is EVEN: function STAYS the same (only sign may change) • Sign: determined by the quadrant the given angle lies in Examples: sin(90° − θ) = cos θ [n=1 odd, Q1, sin>0] cos(90° + θ) = −sin θ [n=1 odd, Q2, cos<0] sin(180° − θ) = sin θ [n=2 even, Q2, sin>0] cos(180° + θ) = −cos θ [n=2 even, Q3, cos<0] sin(270° − θ) = −cos θ [n=3 odd, Q3, sin<0] tan(360° − θ) = −tan θ [n=4 even, Q4, tan<0] sin(−θ) = −sin θ, cos(−θ) = cos θ

The "odd n → change function" rule is the fastest way to handle allied angles. Always determine the quadrant first to get the sign, then apply the function change rule.

2.2

Standard Values — The Complete Table

Memorise this table completely — it appears in almost every trig question

θ	0°	30° (π/6)	45° (π/4)	60° (π/3)	90° (π/2)
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	∞
cosec θ	∞	2	√2	2/√3	1
sec θ	1	2/√3	√2	2	∞
cot θ	∞	√3	1	1/√3	0

📌 Memory Pattern for sin θ (0° to 90°):
sin 0° = √0/2 = 0, sin 30° = √1/2 = 1/2, sin 45° = √2/2 = 1/√2,
sin 60° = √3/2, sin 90° = √4/2 = 1.
Pattern: sin θ = √n/2 where n = 0,1,2,3,4 for θ = 0°,30°,45°,60°,90°. Cos is the reverse (n = 4,3,2,1,0).

📝 TOPIC-WISE PYQ

Ratios & Signs — NDA-Pattern Questions

Q1. The value of sin(150°) + cos(120°) is:

(a) 1 (b) 0 (c) −1 (d) 1/2

Answer: (b) 0
sin 150° = sin(180°−30°) = sin 30° = 1/2.
cos 120° = cos(180°−60°) = −cos 60° = −1/2.
Sum = 1/2 + (−1/2) = 0.

Q2. The value of tan(225°) is:

(a) 1 (b) −1 (c) √3 (d) 1/√3

Answer: (a) 1
225° = 180°+45° → Quadrant III → tan positive.
tan(180°+45°) = tan 45° = 1. (n=2 even, same function.)

Q3. Convert 5π/6 radians to degrees.

(a) 120° (b) 150° (c) 160° (d) 135°

Answer: (b) 150°
5π/6 × (180/π) = 5 × 30 = 150°.

3. Fundamental Identities

3.1

Pythagorean & Reciprocal Identities

Three Pythagorean identities and their derived forms — used in almost every simplification

⚡ All Fundamental Identities

RECIPROCAL IDENTITIES: sin θ ⋅ cosec θ = 1 ⇒ cosec θ = 1/sin θ cos θ ⋅ sec θ = 1 ⇒ sec θ = 1/cos θ tan θ ⋅ cot θ = 1 ⇒ cot θ = 1/tan θ = cos θ/sin θ tan θ = sin θ/cos θ PYTHAGOREAN IDENTITIES (3 base forms): sin²θ + cos²θ = 1 1 + tan²θ = sec²θ ⇒ sec²θ − tan²θ = 1 1 + cot²θ = cosec²θ ⇒ cosec²θ − cot²θ = 1 DERIVED FORMS (express one in terms of another): sin²θ = 1 − cos²θ cos²θ = 1 − sin²θ tan²θ = sec²θ − 1 sec²θ − 1 = tan²θ sin²θ = 1 − cos²θ = (1−cosθ)(1+cosθ)

The three Pythagorean identities come from the same geometric truth (the right triangle) expressed in three different ways. If you know one, you can derive the others by dividing both sides of sin²+cos²=1 by cos² or sin².

Worked Example — Fundamental Identity

Simplify: (sin²θ + cos²θ)(sec²θ − tan²θ)(cosec²θ − cot²θ).

Each factor equals 1: (1)(1)(1) = 1.

This pattern — recognising each factor as 1 from a Pythagorean identity — is a direct NDA question type.

📝 TOPIC-WISE PYQ

Fundamental Identities — NDA-Pattern Questions

Q4. If sin θ = 3/5, find sec θ (given θ is in Q I).

(a) 4/5 (b) 5/4 (c) 3/4 (d) 5/3

Answer: (b) 5/4
cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25 → cosθ = 4/5 (Q I, positive).
sec θ = 1/cos θ = 5/4.

Q5. The value of (1 + tan²θ)(1 + cot²θ) simplifies to:

(a) 1 (b) cosec²θ sec²θ (c) 2 (d) tan²θ cot²θ

Answer: (b) cosec²θ sec²θ
(1+tan²θ)(1+cot²θ) = sec²θ ⋅ cosec²θ = sec²θ cosec²θ.

🔥 TRICKY QUESTIONS

Fundamental Identities — Simplification Traps

🤯 T1. Prove: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ.

LHS = sin²θ+2sinθcosecθ+cosec²θ + cos²θ+2cosθsecθ+sec²θ
= sin²θ+cos²θ + 2(1)+2(1) + cosec²θ+sec²θ
= 1 + 4 + (1+cot²θ)+(1+tan²θ)
= 1+4+1+cot²θ+1+tan²θ = 7+tan²θ+cot²θ ✓
Key: sinθcosecθ=1 and cosθsecθ=1 are the reciprocal products. Always substitute 1 for these immediately.

🤯 T2. If tanθ+cotθ = 2, find tanⁿθ+cotⁿθ for any n.

tanθ+cotθ = 2 → (tanθ+cotθ)² = tan²θ+2tanθcotθ+cot²θ = tan²θ+2(1)+cot²θ = 4.
→ tan²θ+cot²θ = 2.
By induction: tanθ=cotθ=1 (since tan+cot=2 and tan⋅cot=1, they both equal 1).
tanⁿθ+cotⁿθ = 1+1 = 2 for all n.
This is a famous result: if tanθ=cotθ=1 (i.e., θ=45°), then any symmetric sum of powers equals 2.

4. Sum & Difference Formulas

4.1

Compound Angle Identities

These six formulas are the engine of all advanced trig — derive everything else from them

⚡ Sum & Difference Formulas — Complete Set

SIN: sin(A + B) = sin A cos B + cos A sin B sin(A − B) = sin A cos B − cos A sin B COS: cos(A + B) = cos A cos B − sin A sin B cos(A − B) = cos A cos B + sin A sin B TAN: 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 PATTERNS: sin(sum) = sin cos + cos sin [both mixed, same signs] sin(diff) = sin cos − cos sin [both mixed, opposite sign] cos(sum) = cos cos − sin sin [same types, MINUS sign] cos(diff) = cos cos + sin sin [same types, PLUS sign] (cos is the tricky one: sum has minus, difference has plus)

The single biggest error in NDA: writing cos(A+B) = cosA+cosB (wrong!) or sin(A+B) = sinA+sinB (wrong!). These do NOT distribute over addition. The expansion always involves all four terms.

Worked Example — Exact Value Using Sum Formula

Find sin 75°.

75° = 45° + 30°. sin(45°+30°) = sin45°cos30° + cos45°sin30°.

= (1/√2)(√3/2) + (1/√2)(1/2) = √3/(2√2) + 1/(2√2) = (√3+1)/(2√2).

= (√6+√2)/4.

Worked Example — tan(A+B) Application

If tan A = 1/2 and tan B = 1/3, find tan(A+B).

tan(A+B) = (1/2+1/3)/(1−1/2⋅1/3) = (5/6)/(1−1/6) = (5/6)/(5/6) = 1.

Therefore A+B = π/4 = 45°.

📝 TOPIC-WISE PYQ

Sum & Difference Formulas — NDA-Pattern Questions

Q6. cos 15° − sin 15° = ?

(a) √6/2 (b) 1/√2 (c) √3/2 (d) √6/4

Answer: (b) 1/√2
cos15°−sin15° = √2[(1/√2)cos15°−(1/√2)sin15°] = √2[cos45°cos15°−sin45°sin15°]
= √2 cos(45°+15°) = √2 cos60° = √2⋅(1/2) = √2/2 = 1/√2.

Q7. sin(A+B)sin(A−B) = ?

(a) sin²A − sin²B (b) cos²A − cos²B (c) sin²A + sin²B (d) cos²A + sin²B

Answer: (a) sin²A − sin²B
= (sinA cosB + cosA sinB)(sinA cosB − cosA sinB)
= sin²A cos²B − cos²A sin²B = sin²A(1−sin²B) − (1−sin²A)sin²B
= sin²A − sin²A sin²B − sin²B + sin²A sin²B = sin²A − sin²B.

Q8. tan(45°+A) ⋅ tan(45°−A) = ?

(a) −1 (b) 1 (c) tan A (d) 0

Answer: (b) 1
tan(45°+A) = (1+tanA)/(1−tanA). tan(45°−A) = (1−tanA)/(1+tanA).
Product = [(1+tanA)(1−tanA)] / [(1−tanA)(1+tanA)] = 1.

🔥 TRICKY QUESTIONS

Compound Angles — Reverse-Engineering Traps

🤯 T3. Prove: cos²(A−B) − cos²(A+B) = sin 2A sin 2B.

Use a²−b² = (a+b)(a−b) with a=cos(A−B), b=cos(A+B):
= [cos(A−B)+cos(A+B)][cos(A−B)−cos(A+B)].
cos(A−B)+cos(A+B) = (cosAcosB+sinAsinB)+(cosAcosB−sinAsinB) = 2cosAcosB.
cos(A−B)−cos(A+B) = 2sinAsinB.
Product = (2cosAcosB)(2sinAsinB) = 4sinAcosAsinBcosB = sin 2A sin 2B ✓.

🤯 T4. If A+B = π/4, show that (1+tanA)(1+tanB) = 2.

A+B = π/4 → tan(A+B) = 1.
(tanA+tanB)/(1−tanA tanB) = 1 → tanA+tanB = 1−tanA tanB.
Now: (1+tanA)(1+tanB) = 1+tanA+tanB+tanA tanB
= 1+(1−tanA tanB)+tanA tanB [substituting tanA+tanB]
= 1+1 = 2 ✓.
This is a classic NDA result. It generalises: if A+B+C=π/4, then (1+tanA)(1+tanB)(1+tanC)=2√2, etc.

5. Multiple & Sub-Multiple Angle Formulas

5.1

Double Angle, Triple Angle & Half Angle Formulas

cos 2A has THREE equivalent forms — choose based on what appears in the problem

⚡ Double Angle Formulas

sin 2A = 2 sin A cos A = 2 tan A / (1 + tan²A) cos 2A = cos²A − sin²A [form 1 — difference of squares] = 2cos²A − 1 [form 2 — in terms of cos only] = 1 − 2sin²A [form 3 — in terms of sin only] = (1−tan²A)/(1+tan²A) [form 4 — in terms of tan] tan 2A = 2 tan A / (1 − tan²A) Derived results: 1 + cos 2A = 2cos²A ⇒ cos²A = (1+cos2A)/2 1 − cos 2A = 2sin²A ⇒ sin²A = (1−cos2A)/2 sin²A − cos²A = −cos2A sin 2A / (1 + cos 2A) = tan A

The result cos²A = (1+cos2A)/2 and sin²A = (1−cos2A)/2 are used heavily in integration (reducing powers). In NDA, the three forms of cos 2A are tested by asking which form simplifies an expression fastest.

⚡ Triple Angle & Sub-Multiple (Half-Angle) Formulas

TRIPLE ANGLE: sin 3A = 3 sin A − 4 sin³A cos 3A = 4 cos³A − 3 cos A tan 3A = (3tanA − tan³A) / (1 − 3tan²A) SUB-MULTIPLE (HALF ANGLE): sin(A/2) = ±√[(1−cosA)/2] (sign depends on quadrant of A/2) cos(A/2) = ±√[(1+cosA)/2] (sign depends on quadrant of A/2) tan(A/2) = ±√[(1−cosA)/(1+cosA)] = sin A / (1 + cos A) [positive form, no ±] = (1 − cos A) / sin A [alternative positive form] KEY PRODUCT: sin A/2 ⋅ cos A/2 = sin A / 2 (from sin2A = 2sinAcosA with A→A/2)

The two “clean” forms of tan(A/2) without ± signs are very useful: tan(A/2) = sinA/(1+cosA) = (1−cosA)/sinA. These avoid having to determine the sign from the quadrant.

Worked Example — Double Angle

If sin A = 5/13 (A in Q II), find sin 2A.

cos A = −√(1−25/169) = −12/13 (Q II, cos negative).

sin 2A = 2 sin A cos A = 2(5/13)(−12/13) = −120/169.

Worked Example — Triple Angle Factorisation

Express sin 60° using sin 3A formula.

Let A = 20°: sin 60° = 3sin20° − 4sin³20°.

This gives √3/2 = 3sin20° − 4sin³20° — a useful result for exact value derivations.

📝 TOPIC-WISE PYQ

Multiple & Sub-Multiple Angles — NDA-Pattern Questions

Q9. The value of cos²15° − sin²15° is:

(a) 1/2 (b) √3/2 (c) 1 (d) √3/4

Answer: (b) √3/2
cos²15°−sin²15° = cos(2×15°) = cos 30° = √3/2.

Q10. sin 20° sin 40° sin 80° = ?

(a) √3/8 (b) 1/4 (c) √3/4 (d) 3√3/8

Answer: (a) √3/8
Using product formula: sinθ sin(60°−θ) sin(60°+θ) = (1/4)sin3θ.
Here θ=20°: = (1/4)sin60° = (1/4)(√3/2) = √3/8.

Q11. If tan A = 1/2, find the value of sin 2A.

(a) 2/5 (b) 4/5 (c) 3/5 (d) 1/5

Answer: (b) 4/5
sin 2A = 2tanA/(1+tan²A) = 2(1/2)/(1+1/4) = 1/(5/4) = 4/5.

🔥 TRICKY QUESTIONS

Multiple Angles — Classic NDA Traps

🤯 T5. Find the value of cos 36° − cos 72°.

Use identity: cos 36° = (1+√5)/4 and cos 72° = (√5−1)/4 (standard results).
Alternatively: cos72° = 2cos²36°−1. Let c = cos36°.
cos72° = 2c²−1. cos144° = 2cos²72°−1. cos144° = −cos36° = −c.
From 5θ = 180° → cos36°−cos72° = c−(2c²−1) = −2c²+c+1.
Using c = cos36° = (1+√5)/4: result = 1/2.
Direct check: cos36°≈0.809, cos72°≈0.309. Difference ≈ 0.5 = 1/2 ✓.

🤯 T6. Show that sin 3A/sin A − cos 3A/cos A = 2.

LHS = (sin 3A cos A − cos 3A sin A) / (sin A cos A)
= sin(3A−A) / (sin A cos A) [using sin(P−Q) formula numerator]
= sin 2A / (sin A cos A)
= 2 sin A cos A / (sin A cos A) [since sin2A = 2sinAcosA]
= 2 ✓.

6. Transformation Formulas

6.1

Sum-to-Product & Product-to-Sum Formulas

Convert between sums and products to simplify or factorise trigonometric expressions

⚡ Sum-to-Product Formulas (C & D formulas)

If C = A+B and D = A−B, then A = (C+D)/2, B = (C−D)/2: 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] Memory: “Sum/Difference of sines: 2×sin or cos of half-sum × cos or sin of half-diff” sin+sin → 2 sin(half-sum) cos(half-diff) sin−sin → 2 cos(half-sum) sin(half-diff) cos+cos → 2 cos(half-sum) cos(half-diff) cos−cos → −2 sin(half-sum) sin(half-diff) [negative!]

NDA tests these for factorising sin A + sin B type sums and for evaluating things like sin50°+sin10°. Always compute the half-sum and half-difference first.

⚡ Product-to-Sum Formulas

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) [note the ORDER of subtraction: (A−B) first] Shortcut: sin A cos B = (1/2)[sin(A+B) + sin(A−B)] cos A cos B = (1/2)[cos(A+B) + cos(A−B)] sin A sin B = (1/2)[cos(A−B) − cos(A+B)]

Remember for 2sinAsinB: it gives cos(A−B)−cos(A+B) — the subtraction is (A−B) minus (A+B), which is the REVERSE of the order in the cos product formula. This reversal is a very common exam trap.

Worked Example — Sum-to-Product

Simplify (sin 50° + sin 10°) / cos 30°.

sin 50°+sin 10° = 2 sin[(50+10)/2] cos[(50−10)/2] = 2 sin 30° cos 20° = 2(1/2)cos20° = cos20°.

Denominator: cos30° = √3/2.

Result = cos20° / (√3/2). This can be simplified further based on the context.

Worked Example — Product-to-Sum

Express sin 7x cos 3x as a sum.

2 sin 7x cos 3x = sin(7x+3x) + sin(7x−3x) = sin 10x + sin 4x.

sin 7x cos 3x = (sin10x + sin4x)/2.

📝 TOPIC-WISE PYQ

Transformation Formulas — NDA-Pattern Questions

Q12. sin 75° + sin 15° = ?

(a) √3/2 (b) √2 (c) √6/2 (d) 1

Answer: (a) √3/2
sin75°+sin15° = 2sin[(75+15)/2]cos[(75−15)/2] = 2sin45°cos30°
= 2 ⋅ (1/√2) ⋅ (√3/2) = √3/√2 = √3/√2 = √6/2. Option (c).

Q13. cos 20° cos 40° cos 80° = ?

(a) 1/4 (b) √3/8 (c) 1/8 (d) 3/8

Answer: (c) 1/8
Using product: cosθ cos(60°−θ) cos(60°+θ) = (1/4)cos3θ.
Here θ=20°: = (1/4)cos60° = (1/4)(1/2) = 1/8.

Q14. (sin A + sin 3A) / (cos A + cos 3A) = ?

(a) tan 2A (b) cot 2A (c) tan A (d) 2 tan A

Answer: (a) tan 2A
Numerator: sinA+sin3A = 2sin2A cosA. Denominator: cosA+cos3A = 2cos2A cosA.
Ratio = 2sin2A cosA / 2cos2A cosA = tan 2A.

🔥 TRICKY QUESTIONS

Transformation Formulas — Multi-Step NDA Problems

🤯 T7. Prove: sin A + sin 2A + sin 3A = sin 2A(1 + 2cos A).

sinA + sin3A = 2sin[(A+3A)/2]cos[(3A−A)/2] = 2sin2A cosA. (Sum-to-product.)
So: sinA + sin2A + sin3A = 2sin2AcosA + sin2A = sin2A(2cosA+1) = sin2A(1+2cosA) ✓.

🤯 T8. If A+B+C = π, prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

sin2A+sin2B = 2sin(A+B)cos(A−B). Since A+B = π−C → sin(A+B) = sinC.
= 2sinCcos(A−B).
sin2C = 2sinCcosC.
Total = sinC[2cos(A−B)+2cosC] = 2sinC[cos(A−B)+cosC].
Since C = π−A−B: cosC = −cos(A+B).
= 2sinC[cos(A−B)−cos(A+B)] = 2sinC[2sinAsinB] = 4sinAsinBsinC ✓.

📝 Master Formula Sheet — MN17 Trigonometric Ratios & Identities

All critical formulae for rapid pre-exam revision.

◆ Pythagorean Identities

sin²θ+cos²θ = 1
1+tan²θ = sec²θ
1+cot²θ = cosec²θ
sin²θ = (1−cos2θ)/2
cos²θ = (1+cos2θ)/2

➕ Sum & Difference

sin(A±B) = sinAcosB ± cosAsinB
cos(A+B) = cosAcosB − sinAsinB
cos(A−B) = cosAcosB + sinAsinB
tan(A±B) = (tanA±tanB)/(1∓tanAtanB)

✖ Double & Triple Angle

sin2A = 2sinAcosA = 2tanA/(1+tan²A)
cos2A = cos²A−sin²A = 2cos²A−1 = 1−2sin²A
tan2A = 2tanA/(1−tan²A)
sin3A = 3sinA−4sin³A
cos3A = 4cos³A−3cosA

½ Half Angle

sin(A/2) = ±√[(1−cosA)/2]
cos(A/2) = ±√[(1+cosA)/2]
tan(A/2) = sinA/(1+cosA) = (1−cosA)/sinA
1−cos2A = 2sin²A
1+cos2A = 2cos²A

▶ Sum-to-Product

sinC+sinD = 2sin[(C+D)/2]cos[(C−D)/2]
sinC−sinD = 2cos[(C+D)/2]sin[(C−D)/2]
cosC+cosD = 2cos[(C+D)/2]cos[(C−D)/2]
cosC−cosD = −2sin[(C+D)/2]sin[(C−D)/2]

🔗 Product-to-Sum

2sinAcosB = sin(A+B)+sin(A−B)
2cosAcosB = cos(A+B)+cos(A−B)
2sinAsinB = cos(A−B)−cos(A+B)
sinθsin(60°−θ)sin(60°+θ) = (1/4)sin3θ
cosθcos(60°−θ)cos(60°+θ) = (1/4)cos3θ

⚡ Quick Revision Booster — MN17 Trig Ratios & Identities

◆ Quadrant Signs (ASTC)

Q1 (0–90): All positive
Q2 (90–180): Sin positive only
Q3 (180–270): Tan positive only
Q4 (270–360): Cos positive only
Anti-clockwise: A→S→T→C

📈 Standard Values

sin30=1/2, sin45=1/√2, sin60=√3/2
cos: reverse of sin values
tan30=1/√3, tan45=1, tan60=√3
Pattern: sin=√n/2 for n=0,1,2,3,4

➕ Compound Angles

sin(A+B): mixed terms, same signs
cos(A+B): same terms, MINUS sign!
cos(A−B): same terms, PLUS sign
tan(A±B): (tanA±tanB)/(1∓tanAtanB)

✖ Double Angle

sin2A = 2sinAcosA
cos2A: THREE forms (pick best)
cos2A = 1−2sin²A if sin given
cos2A = 2cos²A−1 if cos given
sin²A = (1−cos2A)/2

▶ Sum-to-Product

sinC±sinD: 2 sin/cos × cos/sin
cosC+cosD: 2cos cos
cosC−cosD: −2sin sin (negative!)
Half-sum and half-diff of C and D

🚨 Critical Exam Traps

sin(A+B) ≠ sinA+sinB
cos(A+B) has MINUS: cosAcosB−sinAsinB
cos2A has 3 forms — use the right one
cosC−cosD has a NEGATIVE: −2sin sin
tan(A+B): denominator has 1−tanAtanB
Half-angle signs: check quadrant of A/2

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