Trigonometry (త్రికోణమితి)

cos(A+B)=cosAcosB-sinAsinB. Let A=cos⁻¹x, B=cos⁻¹y → cos(A+B)=xy-√(1-x²)√(1-y²) ✓

Let A=cos⁻¹x, B=cos⁻¹y.
cosA=x, cosB=y.
sinA=√(1-x²), sinB=√(1-y²).
cos(A+B)=cosAcosB-sinAsinB=xy-√(1-x²)√(1-y²).
A+B=cos⁻¹(xy-√(1-x²)√(1-y²)) ✓.

e^(iθ)×e^(-iθ)=e⁰=1=(cosθ+isinθ)(cosθ-isinθ)=cos²θ+sin²θ=1 ✓

e^(iθ) = cosθ+isinθ.
e^(-iθ) = cosθ-isinθ.
Product: e^(iθ)×e^(-iθ) = e⁰ = 1.
(cosθ+isinθ)(cosθ-isinθ) = cos²θ+sin²θ.
∴ cos²θ+sin²θ=1. ✓

Euler's formula unifies trigonometry and complex exponentials!

θ = nπ + (-1)ⁿ sin⁻¹(k), n ∈ ℤ

General solution:
sinθ = k → θ = nπ + (-1)ⁿ α, where α = sin⁻¹(k).

ఉదా: sinθ = 1/2 → α=π/6.
θ = nπ+(-1)ⁿ(π/6).
n=0: π/6 (30°); n=1: π-π/6=5π/6 (150°); n=2: 2π+π/6... ✓

tan2θ = -4/3

tan2θ = 2tanθ/(1-tan²θ) = 2×2/(1-4) = 4/(-3) = -4/3.

Double angle: tan2θ = 2tanθ/(1-tan²θ).
Note: tan²θ=4>1 → 1-tan²θ<0 → tan2θ negative. This means 2θ is in Q2 or Q4!

= π

tan⁻¹1 = π/4.
tan⁻¹2+tan⁻¹3: xy=6>1, so formula gives π+tan⁻¹((2+3)/(1-6))=π+tan⁻¹(-1)=π-π/4=3π/4.
Total = π/4+3π/4 = π.

Standard result: tan⁻¹1+tan⁻¹2+tan⁻¹3=π.

sin 30° = 1/2

Standard values గుర్తు పెట్టుకోండి:
sin 0°=0, sin 30°=1/2, sin 45°=1/√2, sin 60°=√3/2, sin 90°=1.

Trick: 0,30,45,60,90 కు sin values = √0/2, √1/2, √2/2, √3/2, √4/2.
= 0, 1/2, 1/√2, √3/2, 1. ఇలా remember చేయడం easy!

True! sin3θ=3sinθ-4sin³θ is valid for all θ

sin 3θ = 3sinθ - 4sin³θ.
For θ=3°: sin9° = 3sin3°-4sin³3° ✓.

Triple angle formulas:
sin3θ = 3sinθ-4sin³θ.
cos3θ = 4cos³θ-3cosθ.
tan3θ = (3tanθ-tan³θ)/(1-3tan²θ).

cos 30° = √3/2

30°-60°-90° triangle sides: 1, √3, 2.
cos 30° = adj/hyp = √3/2 ≈ 0.866.

గుర్తు: cos 30° = sin 60° = √3/2 (complementary angles!).
Larger angle → larger opposite → larger sin, smaller cos.

= 5π/6

cos(7π/6) = cos(π+π/6) = -cosπ/6 = -√3/2.
cos⁻¹(-√3/2) = 5π/6 (principal value range [0,π]).

cos⁻¹ range = [0,π]. 7π/6 ∉ [0,π].
cos(5π/6) = -√3/2 ✓, and 5π/6 ∈ [0,π] ✓.

sin 60° = √3/2

30-60-90 triangle: sides 1, √3, 2.
sin 60° = opp/hyp = √3/2 ≈ 0.866.

గుర్తు: sin 60° = cos 30° = √3/2.
√3 ≈ 1.732, so √3/2 ≈ 0.866.

= 1/16

cos 108° = cos(180°-72°) = -cos72°.
cos 144° = cos(180°-36°) = -cos36°.
Product = cos36° × cos72° × (-cos72°) × (-cos36°).
= cos²36° × cos²72°.
cos36°=(1+√5)/4×2...
Actually: product = 1/16.

These are roots of Chebyshev polynomial. Standard result: 1/16.

cosec θ = 1/sin θ = hypotenuse/opposite

Reciprocal trigonometric ratios:
cosec θ = 1/sin θ = hyp/opp.
sec θ = 1/cos θ = hyp/adj.
cot θ = 1/tan θ = adj/opp.

గుర్తు: cosec↔sin, sec↔cos, cot↔tan (pairs!).
"co" prefix swaps the ratio!

tan θ = sin θ / cos θ

tan θ = (opp/hyp) ÷ (adj/hyp) = opp/adj = sin θ/cos θ.

గుర్తు: tan = "tangent" — line tangent to unit circle!
Also: cot θ = cos θ/sin θ = 1/tan θ.

cos θ = adjacent/hypotenuse

SOH-CAH-TOA గుర్తు పెట్టుకోండి:
S = Sin = Opposite/Hypotenuse
C = Cos = Adjacent/Hypotenuse
T = Tan = Opposite/Adjacent

"Some Officers Have Curly Auburn Hair Till Old Age"
ఈ sentence వాడి exam లో గుర్తు పెట్టుకోవచ్చు!

Unit circle (r=1) పై x=cosθ, y=sinθ. x²+y²=1 → sin²θ+cos²θ=1 ✓

Unit circle: radius=1, center=origin.
Point P at angle θ from x-axis: P=(cosθ,sinθ).
x-coordinate = cosθ (adjacent/hypotenuse = adj/1 = adj).
y-coordinate = sinθ (opposite/hypotenuse = opp/1 = opp).

This generalizes trig to all quadrants (not just 0°-90°)!
x²+y²=1 → cos²θ+sin²θ=1 directly follows from circle equation.

cos 15° = (√6+√2)/4

cos 15° = cos(45°-30°) = cos45°cos30° + sin45°sin30°.
= (1/√2)(√3/2) + (1/√2)(1/2).
= √3/(2√2) + 1/(2√2) = (√3+1)/(2√2).
= (√3+1)×√2/4 = (√6+√2)/4.

cos 15° = sin 75° = (√6+√2)/4 ≈ 0.966.

cos θ = √(b²-a²)/b

cos²θ = 1-sin²θ = 1-a²/b² = (b²-a²)/b².
cos θ = √(b²-a²)/b.

ఉదా: a=3, b=5 → cos θ=√(25-9)/5=4/5. (3-4-5!)

= 1

sin²30° + cos²30° = (1/2)² + (√3/2)² = 1/4 + 3/4 = 1.

sin²θ + cos²θ = 1 identity verified!
ఇది ఏ angle కైనా = 1. Exam లో ఈ identity remember చేయడం must!

Maximum = √(a²+b²)

asinθ + bcosθ = √(a²+b²) × sin(θ+φ) where tanφ=b/a.
Maximum of sin = 1.
Maximum = √(a²+b²).

ఉదా: 3sinθ+4cosθ max = √(9+16) = 5.
Minimum = -√(a²+b²).

tan 30° = 1/√3

tan 30° = sin 30°/cos 30° = (1/2)/(√3/2) = 1/√3 = √3/3.

గుర్తు: tan values 0,30,45,60,90: 0, 1/√3, 1, √3, ∞.
Pattern: 0 < 1/√3 < 1 < √3 < ∞ (increasing!)

= sinθ + cosθ

sin(90°-θ) = cosθ.
sinθ + sin(90°-θ) = sinθ + cosθ.

ఉదా: θ=30°: sin30°+sin60°=1/2+√3/2=(1+√3)/2.

0

cos 48° = cos 48°.
sin 42° = sin(90°-48°) = cos 48°.
cos 48° - cos 48° = 0.

గుర్తు: sin(90°-θ) = cosθ → if angles sum to 90°, sin = cos!
48° + 42° = 90° → they are complementary!

= 1

Pair up: tan θ × tan(90°-θ) = tanθ × cotθ = 1.
Pairs: (1°,89°), (2°,88°), ..., (44°,46°).
Each pair product = 1. Middle: tan 45° = 1.
Total = 1×1×...×1 = 1.

sin(90°-θ) = cos θ

Complementary angle identities:
sin(90°-θ) = cos θ.
cos(90°-θ) = sin θ.
tan(90°-θ) = cot θ.
cot(90°-θ) = tan θ.

గుర్తు: 90° minus చేస్తే sin↔cos, tan↔cot, sec↔cosec swap అవుతాయి!
ఉదా: sin 30° = cos 60° = 1/2 ✓

cos θ

1-sin²θ = cos²θ (identity).
(1-sin²θ)/cosθ = cos²θ/cosθ = cosθ.

Trigonometric simplifications: Always look for sin²+cos²=1 pattern!

θ = 90°

√3sinθ-cosθ = 2.
Form: R sin(θ-α) = 2. R=√(3+1)=2. sin(θ-α)=1.
R sinθ cosα - R cosθ sinα = 2 where Rcosα=√3, Rsinα=1.
tanα=1/√3 → α=30°.
2sin(θ-30°)=2 → sin(θ-30°)=1 → θ-30°=90° → θ=120°.

Let me verify: √3×sin120°-cos120°=√3×(√3/2)-(-1/2)=3/2+1/2=2 ✓.
θ=120°.

sin(π/6) ≈ π/6 - (π/6)³/6 ≈ 0.5236-0.0239 ≈ 0.4997 ≈ 0.5 ✓

30° = π/6 ≈ 0.5236 radians.
sin x ≈ x - x³/6 = 0.5236 - (0.5236)³/6.
= 0.5236 - 0.1435/6 = 0.5236-0.0239 ≈ 0.4997 ≈ 0.5.

Exact value: sin 30° = 0.5 ✓.
Taylor series gives excellent approximation!

C=(A+B), D=(A-B); sinC+sinD=sin(A+B)+sin(A-B)=2sinAcosB ✓

Let A=(C+D)/2, B=(C-D)/2. Then C=A+B, D=A-B.
sinC+sinD = sin(A+B)+sin(A-B).
= (sinAcosB+cosAsinB)+(sinAcosB-cosAsinB).
= 2sinAcosB = 2sin((C+D)/2)cos((C-D)/2) ✓.

tan 0° = 0; tan 90° = undefined (∞)

tan 0° = sin 0°/cos 0° = 0/1 = 0.
tan 90° = sin 90°/cos 90° = 1/0 = undefined (infinity).

గుర్తు: Division by zero → undefined!
tan 90° ని "infinity" అంటాం — angle vertical అయినపుడు slope infinite.

cos 60° = 1/2

cos values: cos 0°=1, cos 30°=√3/2, cos 45°=1/√2, cos 60°=1/2, cos 90°=0.

గుర్తు: cos values = sin values reverse order!
cos 60° = sin 30° = 1/2.
cos θ = sin(90°-θ) — complementary angles!

sinAcosBcosC + cosAsinBcosC + cosAcosBsinC - sinAsinBsinC

sin(A+B+C) = sin[(A+B)+C].
= sin(A+B)cosC + cos(A+B)sinC.
= (sinAcosB+cosAsinB)cosC + (cosAcosB-sinAsinB)sinC.
= sinAcosBcosC+cosAsinBcosC+cosAcosBsinC-sinAsinBsinC.

cosec θ × sin θ = 1

cosec θ = 1/sin θ.
cosec θ × sin θ = (1/sin θ) × sin θ = 1.

Similarly: sec θ × cos θ = 1, tan θ × cot θ = 1.
గుర్తు: Reciprocal pairs: sin-cosec, cos-sec, tan-cot. Product always = 1!

= 1 (= sin 90°)

sin(A+B) = sinAcosB + cosAsinB.
= sin(30°+60°) = sin 90° = 1.

Or directly: (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1 ✓.

tan θ = 4/3

3sinθ = 4cosθ → sinθ/cosθ = 4/3 → tanθ = 4/3.

గుర్తు: sinθ/cosθ = tanθ directly.
This implies 3-4-5 triangle: sinθ=4/5, cosθ=3/5.

sec²θ - tan²θ = 1

From identity: 1 + tan²θ = sec²θ.
sec²θ - tan²θ = 1.

Similarly: cosec²θ - cot²θ = 1.

sinh x = (eˣ-e⁻ˣ)/2 (hyperbolic); sin x = circular trig ratio

Circular: sin x = opposite/hypotenuse (right triangle).
Hyperbolic: sinh x = (eˣ-e⁻ˣ)/2.

Relations: cosh²x - sinh²x = 1 (vs sin²x+cos²x=1).
sinh(ix) = i sin(x). They're related through complex numbers!
Hyperbolic functions: catenary curve, special relativity uses them.

= 2secθ

cosθ/(1-sinθ) + cosθ/(1+sinθ) = cosθ[(1+sinθ+1-sinθ)/((1-sinθ)(1+sinθ))].
= cosθ × 2/(1-sin²θ) = 2cosθ/cos²θ = 2/cosθ = 2secθ.

-30° లేదా 330°

sin 210° = sin(180°+30°) = -sin30° = -1/2.
sin⁻¹(-1/2) = -30° (principal value, range [-90°,90°]).

గుర్తు: sin⁻¹ range = [-π/2, π/2].
sin⁻¹(sin θ) = θ only if θ ∈ [-90°,90°].
Otherwise use the equivalent angle in range!

1 + tan²θ = sec²θ

Pythagorean identities (3 versions):
1. sin²θ + cos²θ = 1
2. 1 + tan²θ = sec²θ (divide identity 1 by cos²θ)
3. 1 + cot²θ = cosec²θ (divide identity 1 by sin²θ)

గుర్తు: sin²+cos²=1 నుండి మిగతా రెండు derive చేయవచ్చు!

= (1/4)sin3θ

Using product formula:
sinθ × sin(60°-θ) × sin(60°+θ) = (1/4)sin3θ.

Proof: sin(60°-θ)×sin(60°+θ) = sin²60°-sin²θ = 3/4-sin²θ.
sinθ×(3/4-sin²θ) = (3sinθ-4sin³θ)/4 = sin3θ/4.

Similarly: cosθcos(60°-θ)cos(60°+θ) = (1/4)cos3θ.

No quadrant — at least one is always positive!

ASTC rule (All Students Take Calculus):
Q1: All positive (A).
Q2: Sin positive (S).
Q3: Tan positive (T).
Q4: Cos positive (C).

In every quadrant, at least one ratio is positive.
No quadrant has ALL negative ratios!

1 + 2 sin θ cos θ = 1 + sin 2θ

(sin θ+cos θ)² = sin²θ+2sinθcosθ+cos²θ.
= (sin²θ+cos²θ) + 2sinθcosθ.
= 1 + 2sinθcosθ.
= 1 + sin 2θ.

ఉదా: θ=45° → (1/√2+1/√2)²=(√2)²=2=1+2×(1/2)=2 ✓

sin θ = 1/√2

tanθ + cotθ = sinθ/cosθ + cosθ/sinθ = (sin²θ+cos²θ)/(sinθcosθ) = 1/(sinθcosθ) = 2.
sinθcosθ = 1/2.
sin2θ = 2sinθcosθ = 1 → 2θ = 90° → θ = 45°.
sinθ = sin45° = 1/√2.

θ = nπ లేదా θ = (2n+1)π/4

sinθ+sin3θ = 2sin2θcosθ.
So: 2sin2θcosθ+sin2θ = 0.
sin2θ(2cosθ+1) = 0.
sin2θ=0 → 2θ=nπ → θ=nπ/2.
2cosθ+1=0 → cosθ=-1/2 → θ=2nπ±2π/3.

Combining: θ=nπ/2 or θ=2nπ±2π/3.

sec 60° = 2

sec θ = 1/cos θ.
sec 60° = 1/cos 60° = 1/(1/2) = 2.

గుర్తు: sec values: 1, 2/√3, √2, 2, undefined (0°,30°,45°,60°,90°).
sec 60° = 2 అంటే hypotenuse = 2 × adjacent at 60°!

tan 15° = 2-√3

tan 15° = tan(45°-30°) = (tan45°-tan30°)/(1+tan45°tan30°).
= (1-1/√3)/(1+1/√3) = (√3-1)/(√3+1).
Rationalize: (√3-1)²/((√3)²-1²) = (3-2√3+1)/2 = (4-2√3)/2 = 2-√3.

tan 15° ≈ 0.268 = 2-1.732 = 0.268 ✓

cot 45° = 1

cot θ = 1/tan θ = cos θ/sin θ.
cot 45° = 1/tan 45° = 1/1 = 1.

గుర్తు: tan 45°=1 → cot 45°=1 (reciprocal of 1 = 1).
tan మరియు cot ఒకే value కలిగిన angle = 45°!

= cos180°+isin180° = -1

(cos30°+isin30°)⁶ = cos(6×30°)+isin(6×30°).
= cos180°+isin180° = -1+0i = -1.

De Moivre's theorem: (cosθ+isinθ)ⁿ = cosnθ+isinnθ.
This connects trigonometry and complex numbers beautifully!

= 1/8

sin70°=cos20°. Formula: sinθ×sin(60°-θ)×sin(60°+θ)=(1/4)sin3θ.
θ=10°: sin10°×sin50°×sin70°.
sin(60°-10°)=sin50°, sin(60°+10°)=sin70° ✓.
= (1/4)sin30° = (1/4)(1/2) = 1/8.

Same as cos product formula but for sin!

LHS = sinθ/cosθ + cosθ/sinθ = 1/(sinθcosθ) = secθ×cosecθ = RHS ✓

LHS = tanθ+cotθ = sinθ/cosθ + cosθ/sinθ.
= (sin²θ+cos²θ)/(sinθcosθ) = 1/(sinθcosθ).
= (1/cosθ)(1/sinθ) = secθ × cosecθ = RHS ✓