Trigonometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision.

Trigonometry — Key Facts for CUET • Essential formula: Pythagorean identity — sin²θ + cos²θ = 1 (valid for all θ; all other trig identities derive from it) • Most tested CUET concept: Evaluating sin, cos, tan of standard angles (0°, 30°, 45°, 60°, 90°) using the unit circle or special triangles — a frequent MCQ type. • Common mistake: Forgetting quadrant sign rules when solving trig equations — e.g., sinθ = ½ gives θ = 30° or 150° in the 1st and 2nd quadrants respectively; missing the second solution means losing half the marks. • Key technique: Use complementary angle identities (sinθ = cos(90°−θ), tanθ = sinθ/cosθ) to convert unfamiliar angles into standard ones before evaluating. • Important exception: At θ = 90°, cosθ = 0, so tanθ and secθ are undefined — many CUET options trap you with a finite value for tan 90°. • Most frequent question type: Simplifying or evaluating an expression involving sin 15°, cos 75°, tan 15°, etc., using half-angle or sum/difference formulas. ⚡ Exam tip: For any MCQ involving sin(θ ± φ) or cos(θ ± φ), always expand using the sum/difference formula first — do not substitute numeric values before expanding, or you risk applying the wrong sign convention.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Trigonometry — CUET Study Guide

Trigonometric ratios — sin, cos, tan and their reciprocals (cosec, sec, cot) — relate the angles of a right triangle to the ratios of its sides. For an acute angle θ: sinθ = opposite/hypotenuse, cosθ = adjacent/hypotenuse, tanθ = opposite/adjacent. These extend to all real angles via the unit circle: a point (cosθ, sinθ) on the circle of radius 1.

Fundamental identities:

• Pythagorean: sin²θ + cos²θ = 1
• Quotient: tanθ = sinθ/cosθ
• Reciprocal: cosecθ = 1/sinθ, secθ = 1/cosθ, cotθ = 1/tanθ

Sum and difference formulas (frequently tested in MCQs):

• sin(A ± B) = sinA cosB ± cosA sinB
• cos(A ± B) = cosA cosB ∓ sinA sinB
• tan(A ± B) = (tanA ± tanB)/(1 ∓ tanA tanB)

Double-angle and half-angle:

• sin2θ = 2sinθ cosθ
• cos2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ
• sin(θ/2) = ±√((1−cosθ)/2); cos(θ/2) = ±√((1+cosθ)/2) — sign determined by quadrant of θ/2.

Solved Example 1: Evaluate sin 15° without a calculator.

• sin15° = sin(45°−30°) = sin45°cos30° − cos45°sin30° = (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4 ≈ (√6 − √2)/4

Solved Example 2: Solve for θ (0° ≤ θ < 360°): 2cos²θ − 1 = 0.

• Using cos2θ identity: 2cos²θ − 1 = cos2θ = 0 → 2θ = 90° or 270° → θ = 45° or 135°. But also cos²θ = ½ → cosθ = ±1/√2 → θ = 45°, 135°, 225°, 315°. ✓ Both methods confirm four solutions in 0–360°.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer timeline.

Trigonometry — Comprehensive CUET Notes

Extension to all real angles: By placing a ray rotating from the positive x-axis, any real angle θ maps to a point (cosθ, sinθ) on the unit circle. This extends trigonometry beyond acute triangles to all real numbers. The functions become periodic: sin(θ + 2π) = sinθ, cos(θ + 2π) = cosθ, tan(θ + π) = tanθ. This periodicity underpins Fourier analysis and wave modelling.

Inverse trigonometric functions — arcsin, arccos, arctan — map from [−1, 1] (or R for arctan) back to their principal value ranges: arcsin: [−π/2, π/2], arccos: [0, π], arctan: (−π/2, π/2). Key identities include arcsin(sinθ) ≠ θ unless θ lies in [−π/2, π/2]; similarly for arccos and arctan. These range restrictions are a common trap in CUET inverse trig problems.

Transformation identities (product-to-sum and sum-to-product):

• sinA cosB = ½[sin(A+B) + sin(A−B)]
• cosA cosB = ½[cos(A−B) + cos(A+B)]
• sinA + sinB = 2sin((A+B)/2) cos((A−B)/2)

Solving general trig equations: For equations of the form sin nθ = k (|k| ≤ 1), the general solution is nθ = arcsin(k) + 2πm or nθ = π − arcsin(k) + 2πm, giving θ = [arcsin(k) + 2πm]/n and [π − arcsin(k) + 2πm]/n. Always check the domain restriction specified in the question.

Advanced solved example: Solve tan³θ + 3tanθ = 0 for θ in (0°, 90°).

• tanθ(tan²θ + 3) = 0 → tanθ = 0 or tan²θ = −3 (impossible for real θ)
• tanθ = 0 → θ = 0°, 180° … → in (0°, 90°), only θ = 0°? But 0° is at the boundary. Actually tanθ = 0 also gives θ = 0° (excluded) and θ = 180° etc. So in (0°, 90°), there is no solution — this is the subtlety of open intervals. However many textbooks include 0° as a trivial solution; verify domain: if (0°, 90°) strictly, then tanθ = 0 gives no solution. Check alternatives: Could factor tanθ(tanθ + √3)(tanθ − √3) = 0, giving tanθ = 0, ±√3. tanθ = √3 → θ = 60° ✓. tanθ = −√3 is negative, not in (0°, 90°). So the answer is θ = 60°.

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