Trigonometric Functions 📐

Class 10 mein right triangle mein sin/cos/tan seekha tha — ab Class 11 mein yeh functions poore number line par extend ho jaate hain!

🌊 Waves aur Cycles

Socho ek wheel ek circle par ghoom raha hai. Agar us wheel ki height ko time ke saath plot karo, toh sine wave banta hai! Yahi reason hai ki trigonometric functions physics, music, aur engineering mein everywhere hain. Class 10 mein humne sirf 0°–90° ke angles ke liye trig seekha tha. Ab Class 11 mein hum saare angles ke liye — negative angles bhi, 360° se bade angles bhi — trig extend karte hain. Iske liye Unit Circle ka concept use karte hain.

1. Degree se Radian — kyun aur kaise?

Degree ek arbitrary unit hai (360 kyon? Babylonians ne choose kiya tha!). Radian ek natural unit hai: 1 radian = angle jab arc length = radius. Yahi reason hai ki advanced maths mein hamesha radians use hote hain — formulas simpler hote hain.

π radians = 180° ⟹ 1° = π/180 rad | 1 rad = 180°/π ≈ 57.3°

Conversion trick: Degree → Radian: ×(π/180) | Radian → Degree: ×(180/π)

DegreeRadianDegreeRadian
0180°π
30°π/6270°3π/2
45°π/4360°
60°π/3−90°−π/2
90°π/2−180°−π

2. Unit Circle — Trig Functions ka Asli Matlab

Unit circle ka radius = 1 hota hai aur center origin par. Kisi bhi angle θ ke liye, circle par point P = (cos θ, sin θ) hota hai. Yahi definition hai Class 11 mein — right triangle nahi, unit circle!

P(cos θ, sin θ) sin θ cos θ θ 1 1 Q1 Q2 Q3 Q4

Unit Circle: radius = 1. Koi bhi angle θ ke liye P = (cos θ, sin θ)

3. 6 Trigonometric Ratios

P = (x, y) unit circle par hai, angle θ ke liye — toh 6 trig functions yahan se define hote hain:

sin θ

y/r = y

"Perpendicular/Hypotenuse" — y-coordinate

cos θ

x/r = x

"Base/Hypotenuse" — x-coordinate

tan θ

y/x = sin/cos

"P/B" — slope of radius

cosec θ

1/sin θ

sin ka reciprocal. Defined jab sin≠0

sec θ

1/cos θ

cos ka reciprocal. Defined jab cos≠0

cot θ

1/tan θ = cos/sin

tan ka reciprocal. Defined jab tan≠0

4. Standard Values Table (Yaad karo!)

θsin θcos θtan θcosec θsec θcot θ
0101
30° (π/6)1/2√3/21/√322/√3√3
45° (π/4)1/√21/√21√2√21
60° (π/3)√3/21/2√32/√321/√3
90° (π/2)1010
180° (π)0−10−1

✋ "All Silver Tea Cups" — Quadrant Sign Trick

Har quadrant mein kaunse trig functions positive hain — yeh yaad karne ka trick hai "All Silver Tea Cups":

Q1: All (sin, cos, tan sab positive) | Q2: Sin positive (baaki negative) | Q3: Tan positive (baaki negative) | Q4: Cos positive (baaki negative)

5. Pythagorean Identities

Identity 1 (sin²+cos²)

sin²θ + cos²θ = 1

Unit circle se directly — x²+y²=1. Sabse important identity!

Identity 2 (tan²+1)

1 + tan²θ = sec²θ

Identity 1 ko cos²θ se divide karo.

Identity 3 (cot²+1)

1 + cot²θ = cosec²θ

Identity 1 ko sin²θ se divide karo.

6. Sum aur Difference Formulas

Jab two angles ka sum ya difference ka trig value chahiye — yeh formulas kaam aate hain:

sin(A+B) = sinA·cosB + cosA·sinB
sin(A−B) = sinA·cosB − cosA·sinB
cos(A+B) = cosA·cosB − sinA·sinB
cos(A−B) = cosA·cosB + sinA·sinB
tan(A+B) = (tanA + tanB)/(1 − tanA·tanB)

✅ sin 75° nikalo (75 = 45 + 30)

sin 75° = 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   (rationalize karke)

7. Double Angle Formulas

sin 2A = 2 sinA cosA
cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A
tan 2A = 2tanA / (1 − tan²A)

Trick: A+A = 2A — sum formula mein A=B rakh do!

8. General Solutions

Trig equations ke infinite solutions hote hain (kyunki sin/cos periodic hain with period 2π). General solution batata hai saare possible values:

sin θ = 0 ⟹ θ = nπ, n ∈ ℤ
cos θ = 0 ⟹ θ = (2n+1)π/2, n ∈ ℤ
tan θ = 0 ⟹ θ = nπ, n ∈ ℤ
sin θ = sin α ⟹ θ = nπ + (−1)ⁿα, n ∈ ℤ
cos θ = cos α ⟹ θ = 2nπ ± α, n ∈ ℤ

9. Practice Questions

Q1. 210° ko radians mein convert karo.

Solution

210 × π/180 = 7π/6 radians

Q2. Prove karo: (sin θ + cosec θ)² = sin²θ + cosec²θ + 2

Solution

LHS = sin²θ + 2sinθ·cosecθ + cosec²θ = sin²θ + 2sinθ·(1/sinθ) + cosec²θ = sin²θ + 2 + cosec²θ = RHS ✓

Q3. cos 15° ki value nikalo using cos(45°−30°).

Solution

cos 15° = cos45°cos30° + sin45°sin30° = (1/√2)(√3/2)+(1/√2)(1/2) = (√3+1)/(2√2) = (√6+√2)/4

Q4. tan²θ + 1 = sec²θ prove karo.

Solution

sin²θ + cos²θ = 1 ko cos²θ se divide karo → tan²θ + 1 = sec²θ ✓

Q5. sin θ = 1/2 ka general solution likho.

Solution

sin θ = sin(π/6) → θ = nπ + (−1)ⁿ(π/6), n ∈ ℤ

← Relations & Functions 🧪 Unit Circle Lab →