Trigonometry (త్రికోణమితి)

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 ✓

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) ✓.

θ = 45°

tan 45° = 1.
tan θ = 1 → θ = 45° (in first quadrant).

గుర్తు: tan 45° = 1 — most important value!
Slope of 45° line = 1 (equal rise and run).

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).

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°!

sin θ=5/13, cos θ=12/13, tan θ=5/12

SOH-CAH-TOA:
sin θ = opp/hyp = 5/13.
cos θ = adj/hyp = 12/13.
tan θ = opp/adj = 5/12.

Verify: sin²+cos²=25/169+144/169=169/169=1 ✓
5-12-13 is a Pythagorean triple — exam favourite!

= sinθ + cosθ

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

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

cos(A-B)=cosAcosB+sinAsinB; cos(A+B)=cosAcosB-sinAsinB. Subtract→2sinAsinB=cos(A-B)-cos(A+B) ✓

RHS = ½[cos(A-B)-cos(A+B)].
cos(A-B) = cosAcosB+sinAsinB.
cos(A+B) = cosAcosB-sinAsinB.
Difference = 2sinAsinB.
½×2sinAsinB = sinAsinB = LHS ✓.

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!

sin 90° = 1 మరియు cos 0° = 1

sin 90° = 1 (maximum value of sin).
cos 0° = 1 (maximum value of cos).

గుర్తు: sin మరియు cos values ఎల్లప్పుడూ -1 మరియు +1 మధ్య ఉంటాయి.
Maximum = 1 (sin 90°, cos 0°), Minimum = -1 (sin 270°, cos 180°).

Modulus = r, Argument = θ

z = r(cosθ+isinθ) = re^(iθ) (Euler's form).
|z| = r (modulus = distance from origin).
arg(z) = θ (argument = angle from positive x-axis).

Polar form: (r, θ). Multiplication: r₁r₂, θ₁+θ₂.
This connects algebra and trigonometry elegantly!

C=π-A-B. Use double angle, sum formulas step by step

sin²A+sin²B+sin²C.
= (1-cos2A)/2+(1-cos2B)/2+(1-cos2C)/2.
= 3/2 - (cos2A+cos2B+cos2C)/2.
cos2A+cos2B+cos2C = -1-4cosAcosBcosC (known identity for A+B+C=π).
= 3/2-(-1-4cosAcosBcosC)/2 = 3/2+1/2+2cosAcosBcosC = 2+2cosAcosBcosC.
= 2(1+cosAcosBcosC) ✓.

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.

= 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!

cos 90° = 0

cos 2×45° = 1-2sin²45° = 1-2×(1/2) = 1-1 = 0.
cos 90° = 0 ✓.

Double angle formula: cos 2θ = 1-2sin²θ = 2cos²θ-1 = cos²θ-sin²θ.
All three are equivalent!

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!)

= √3/4

cos²A-sin²B is not a standard identity directly.
Use: cos²45°-sin²15° = (1+cos90°)/2 - (1-cos30°)/2.
= (1+0)/2 - (1-√3/2)/2.
= 1/2 - 1/2 + √3/4 = √3/4.

Or: cos²A-sin²B = cos(A+B)cos(A-B).
= cos(45°+15°)cos(45°-15°) = cos60°cos30° = (1/2)(√3/2) = √3/4.

cos θ

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

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

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 ✓

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 ✓

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).

tan 60° = √3

tan 60° = sin 60°/cos 60° = (√3/2)/(1/2) = √3.

గుర్తు: tan 60° = cot 30° = √3.
√3 ≈ 1.732 — greater than 1 (angle > 45°).

= 1

a³+b³ = (a+b)³-3ab(a+b) where a=sin²θ, b=cos²θ, a+b=1.
sin⁶θ+cos⁶θ = 1-3sin²θcos²θ.
sin⁶θ+cos⁶θ+3sin²θcos²θ = 1.

ఉదా: θ=45°: (1/2)³+(1/2)³+3×(1/2)(1/2)=1/8+1/8+3/4=1 ✓

T₂(cosθ) = 2cos²θ-1 = cos2θ ✓

cos2θ = 2cos²θ-1.
Let x=cosθ: T₂(x) = 2x²-1.
T₂(cosθ) = 2cos²θ-1 = cos2θ ✓.

Chebyshev polynomials: T_n(cosθ) = cos(nθ).
T₁(x)=x, T₂(x)=2x²-1, T₃(x)=4x³-3x...

sin20°sin40°sin80° = (1/4)sin60° = √3/8

sinα×sin(60°+α)×sin(60°-α) = (1/4)sin3α.
α=20°: sin20°×sin80°×sin40° = (1/4)sin60° = (1/4)(√3/2) = √3/8.

sin60°+20°=sin80°, sin60°-20°=sin40° ✓.
Standard result: sin20°sin40°sin80° = √3/8!

sin 45° = 1/√2 = √2/2

45°-45°-90° triangle లో legs equal.
sin 45° = opp/hyp = 1/√2 = √2/2 ≈ 0.707.

గుర్తు: 1/√2 rationalize చేస్తే √2/2.
sin 45° = cos 45° = 1/√2 (ఇవి equal — 45° symmetric!)

= 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.

sin²θ + cos²θ = 1

ఇది Trigonometry లో అత్యంత ముఖ్యమైన identity!
Proof: Right triangle లో (opp² + adj²)/hyp² = hyp²/hyp² = 1.
sin²θ + cos²θ = 1 ఎల్లప్పుడూ valid — ఏ angle అయినా.

ఇందు నుండి: sin²θ = 1-cos²θ, cos²θ = 1-sin²θ.

-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!

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°.

θ = 0° లేదా θ = 60°

Let c=cosθ: 2c²-3c+1=0 → (2c-1)(c-1)=0.
c=1/2 → cosθ=1/2 → θ=60°.
c=1 → cosθ=1 → θ=0°.

గుర్తు: Trig equations → quadratic గా treat → factor చేయాలి!

cos A = √((m²-n²)/(m²-1))... let me derive

sinA = n sinB → sinB = sinA/n.
tanA = m tanB → sinA/cosA = m sinB/cosB → cosB = m sinB cosA/sinA = m cosA/n.
sin²B+cos²B=1: sin²A/n²+m²cos²A/n²=1.
sin²A+m²cos²A=n².
1-cos²A+m²cos²A=n² → (m²-1)cos²A=n²-1.
cosA=√((n²-1)/(m²-1)).

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!

= 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] ✓.

= 2

sinθ + 1/sinθ = 2 → sin²θ - 2sinθ + 1 = 0 → (sinθ-1)² = 0 → sinθ = 1.
sin⁹θ = 1⁹ = 1. cosec⁹θ = 1.
Sum = 2.

గుర్తు: sinθ+cosecθ=2 → sinθ=1 → θ=90° (only solution).

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.

= 1/8

Formula: cosθ×cos(60°-θ)×cos(60°+θ) = (1/4)cos3θ.
θ=20°: cos20°×cos40°×cos80°.
cos(60°-20°)=cos40°, cos(60°+20°)=cos80° ✓.
= (1/4)cos60° = (1/4)(1/2) = 1/8.

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.

Opposite = 6

sin θ = opp/hyp = 3/5.
opp = (3/5) × 10 = 6.

Check: adj=√(100-36)=8. sin θ=6/10=3/5 ✓, cos θ=8/10=4/5.
6-8-10 = 3-4-5 triple!

Let P=A+B, Q=A-B. cos(A+B)+cos(A-B)=2cosAcosB → P,Q substitution ✓

cos(A+B) = cosAcosB-sinAsinB.
cos(A-B) = cosAcosB+sinAsinB.
Sum = 2cosAcosB.
Let P=A+B, Q=A-B → A=(P+Q)/2, B=(P-Q)/2.
cosP+cosQ = 2cos((P+Q)/2)cos((P-Q)/2). ✓

= 1 - 2sinθcosθ = 1 - sin2θ

(sinθ-cosθ)² = sin²θ - 2sinθcosθ + cos²θ.
= (sin²θ+cos²θ) - 2sinθcosθ.
= 1 - sin2θ.

Compare: (sinθ+cosθ)² = 1+sin2θ.
Difference = 4sinθcosθ = 2sin2θ.

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.

cosec 30° = 2

cosec θ = 1/sin θ.
cosec 30° = 1/sin 30° = 1/(1/2) = 2.

గుర్తు: cosec values are reciprocals of sin values.
sin 30°=1/2 → cosec 30°=2. Easy!

= 2

sin 18° × cosec 18° = sin 18° × (1/sin 18°) = 1.
cos 72° × sec 72° = cos 72° × (1/cos 72°) = 1.
Total = 1+1 = 2.

గుర్తు: sin×cosec = cos×sec = tan×cot = 1 (reciprocal pairs)!

= 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!

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!

sin 75° = (√6+√2)/4

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) = (√3+1)√2/4 = (√6+√2)/4.

sin 75° ≈ 0.966.

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!

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.