🎯 Key Points
- Pythagorean identities: sin²θ+cos²θ=1, 1+tan²θ=sec²θ, 1+cot²θ=cosec²θ — all three derive from the same unit-circle relation, just divided by cos²θ or sin²θ
- Compound angles: sin(A±B)=sinAcosB±cosAsinB, cos(A±B)=cosAcosB∓sinAsinB — note cos flips the sign (∓), a common error source
- Double angle cos2A has THREE equivalent forms (cos²A-sin²A = 1-2sin²A = 2cos²A-1) — pick whichever form eliminates the variable you don't need
- Sine rule a/sinA=b/sinB=c/sinC=2R relates sides to OPPOSITE angles; Cosine rule a²=b²+c²-2bc·cosA is the generalized Pythagoras (reduces to it when A=90°)
Trigonometry
Trigonometry deals with relationships between angles and sides of triangles.
Trigonometric Ratios
Unit circle with standard angles 0, 30, 45, 60, 90 degrees; at 45 degrees the point is (cos45, sin45) = (1/sqrt2, 1/sqrt2).
- sin θ = opposite/hypotenuse
- cos θ = adjacent/hypotenuse
- tan θ = opposite/adjacent = sin/cos
- cosec θ = 1/sin θ | sec θ = 1/cos θ | cot θ = 1/tan θ
Standard Values
| θ | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | ∞ |
Pythagorean Identities
- sin²θ + cos²θ = 1
- 1 + tan²θ = sec²θ
- 1 + cot²θ = cosec²θ
Compound Angles
- sin(A±B) = sinA cosB ± cosA sinB
- cos(A±B) = cosA cosB ∓ sinA sinB
- tan(A+B) = (tanA + tanB)/(1 - tanA tanB)
Double and Half Angles
- sin2A = 2 sinA cosA
- cos2A = cos²A - sin²A = 1 - 2sin²A = 2cos²A - 1
- tan2A = 2tanA/(1-tan²A)
Sine and Cosine Rules
- Sine rule: a/sinA = b/sinB = c/sinC = 2R
- Cosine rule: a² = b² + c² - 2bc cosA
Signs of Ratios in Quadrants (ASTC)
The rule "All Students Take Coffee" tells which ratios are positive in each quadrant:
| Quadrant | Angle range | Positive ratios |
|---|---|---|
| I | 0°–90° | All |
| II | 90°–180° | Sin (& cosec) |
| III | 180°–270° | Tan (& cot) |
| IV | 270°–360° | Cos (& sec) |
Allied Angles (Reduction Formulae)
- Odd multiples of 90° (90°, 270°) flip the ratio: sin ↔ cos, tan ↔ cot, sec ↔ cosec.
- Even multiples of 90° (180°, 360°) keep the ratio same; only the sign changes per quadrant (ASTC).
- sin(90° − θ) = cosθ, cos(90° − θ) = sinθ, tan(90° − θ) = cotθ (complementary angles).
- sin(180° − θ) = sinθ, cos(180° − θ) = −cosθ; sin(−θ) = −sinθ (odd), cos(−θ) = cosθ (even).
- sin(360° + θ) = sinθ — all six ratios are periodic, so adding full turns leaves them unchanged.
Sub-Multiple (Half) Angle Formulae
- sinA = 2 sin(A/2) cos(A/2); cosA = 1 − 2sin²(A/2) = 2cos²(A/2) − 1.
- sin²(A/2) = (1 − cosA)/2; cos²(A/2) = (1 + cosA)/2; tan²(A/2) = (1 − cosA)/(1 + cosA).
- Weierstrass (t = tan(A/2)): sinA = 2t/(1 + t²), cosA = (1 − t²)/(1 + t²), tanA = 2t/(1 − t²).
Product ↔ Sum Transformations
- 2 sinA cosB = sin(A+B) + sin(A−B); 2 cosA sinB = sin(A+B) − sin(A−B).
- 2 cosA cosB = cos(A+B) + cos(A−B); 2 sinA sinB = cos(A−B) − cos(A+B).
- sinC + sinD = 2 sin((C+D)/2) cos((C−D)/2); sinC − sinD = 2 cos((C+D)/2) sin((C−D)/2).
- cosC + cosD = 2 cos((C+D)/2) cos((C−D)/2); cosC − cosD = −2 sin((C+D)/2) sin((C−D)/2).
General Solutions of Trigonometric Equations
| Equation | General solution (n ∈ Z) |
|---|---|
| sinθ = 0 | θ = nπ |
| cosθ = 0 | θ = (2n + 1)π/2 |
| tanθ = 0 | θ = nπ |
| sinθ = sinα | θ = nπ + (−1)ⁿ α |
| cosθ = cosα | θ = 2nπ ± α |
| tanθ = tanα | θ = nπ + α |
For sin²θ = sin²α (also cos²θ = cos²α or tan²θ = tan²α), the general solution is θ = nπ ± α.
Graphs and Periodicity
- sinx and cosx have period 2π and range [−1, 1]; cosx = sin(x + π/2) is just sinx shifted left by π/2.
- tanx and cotx have period π and range (−∞, ∞); tanx has vertical asymptotes at x = (2n+1)π/2.
- secx and cosecx have period 2π and range (−∞, −1] ∪ [1, ∞).
- Period of sin(kx) or cos(kx) is 2π/|k|; period of tan(kx) is π/|k|; |sinx| has period π.
- sinx is an odd function (graph symmetric about origin); cosx is even (symmetric about the y-axis).
🚀 JEE Advanced Edge
General solution vs principal solution: sinθ=1/2 has principal solutions θ=30° or 150° within [0°,360°), but the GENERAL solution accounting for periodicity is θ=nπ+(-1)ⁿ(π/6) for sin, or θ=nπ±π/3 for cos-type equations — JEE problems on trig equations almost always want the general solution unless explicitly restricted to a range, and using the wrong general-solution form (treating sin like cos) is a frequent error.
Conditional trigonometric identities in a triangle: Since A+B+C=π in any triangle, identities like tanA+tanB+tanC = tanA·tanB·tanC hold ONLY because of this angle-sum constraint — recognizing when a problem implicitly uses A+B+C=π (rather than treating A, B, C as independent angles) is key to simplifying triangle-based trig expressions efficiently.
Worked problem: If A+B = 45°, find the value of (1+tanA)(1+tanB). Approach: Since tan(A+B) = (tanA+tanB)/(1-tanA·tanB) = tan45° = 1, we get tanA+tanB = 1-tanA·tanB, i.e. tanA+tanB+tanA·tanB = 1. Now expand (1+tanA)(1+tanB) = 1+tanA+tanB+tanA·tanB = 1+1 = 2.
Worked Example: Solving a Trigonometric Equation
Solve 2sin²x + sinx − 1 = 0 for x ∈ [0, 2π].
Let s = sinx. Then 2s² + s − 1 = 0 factors as (2s − 1)(s + 1) = 0, giving s = ½ or s = −1.
sinx = ½ → x = π/6 or x = 5π/6. sinx = −1 → x = 3π/2. Solutions: x = π/6, 5π/6, 3π/2. Always check that each solution lies within the required interval.
Worked Example: Finding sin θ and cos θ from tan θ
If tan θ = 3/4 and θ is in the third quadrant, find sin θ and cos θ.
Use sec²θ = 1 + tan²θ = 1 + 9/16 = 25/16, so cos θ = ±4/5. In Q3, cos θ is negative → cos θ = −4/5.
Then sin θ = tan θ · cos θ = (3/4)(−4/5) = −3/5. Answer: sin θ = −3/5, cos θ = −4/5. Always fix the sign using the quadrant, not just the magnitude.