Trigonometry (త్రికోణమితి)

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] ✓.

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)) ✓.

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 ✓

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.

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).

= 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 ✓

cos θ = 3/5, tan θ = 4/3

cos²θ = 1-sin²θ = 1-16/25 = 9/25 → cos θ = 3/5.
tan θ = sin θ/cos θ = (4/5)/(3/5) = 4/3.

3-4-5 triangle: opp=4, adj=3, hyp=5 ✓.

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) ✓.

= 1

(a-b)(a+b) = a²-b² formula.
sec²θ - tan²θ = 1 (identity).

గుర్తు: sec²θ-tan²θ=1 → (secθ-tanθ)(secθ+tanθ)=1.
ఇది reciprocal property: (secθ-tanθ) = 1/(secθ+tanθ)!

tan(A+B) = C/(1-D)

tan(A+B) = (tanA+tanB)/(1-tanA×tanB) = C/(1-D).

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 ✓.

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 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.

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.

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°!

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!

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.

sec²θ - tan²θ = 1

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

Similarly: cosec²θ - cot²θ = 1.

LHS = sin²θ/cos²θ - sin²θ = sin²θ(1-cos²θ)/cos²θ = sin²θ×sin²θ/cos²θ = tan²θ×sin²θ = RHS ✓

LHS = tan²θ - sin²θ.
= sin²θ/cos²θ - sin²θ.
= sin²θ(1/cos²θ - 1).
= sin²θ × (1-cos²θ)/cos²θ.
= sin²θ × sin²θ/cos²θ.
= sin²θ × tan²θ = RHS ✓

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 + 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 ✓

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.

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.

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!

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!)

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)).

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 ✓

= (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.

= 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.

(1+tanθ)/(1-tanθ)

Divide numerator and denominator by cosθ:
(cosθ+sinθ)/cosθ = 1+tanθ.
(cosθ-sinθ)/cosθ = 1-tanθ.
Result = (1+tanθ)/(1-tanθ) = tan(45°+θ).

ఉదా: θ=0°: (1+0)/(1-0)=1. θ=45°: undefined (tan90°).

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!

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!

sin 60° = 2×(1/2)×(√3/2) = √3/2 ✓

LHS: sin 2×30° = sin 60° = √3/2.
RHS: 2×sin 30°×cos 30° = 2×(1/2)×(√3/2) = √3/2.
LHS = RHS ✓.

sin 2θ = 2sinθcosθ — Double Angle Formula!

sin θ=5/13, cos θ=12/13

tan=opp/adj=5/12 → opp=5,adj=12.
hyp=√(25+144)=√169=13.
sin=5/13, cos=12/13.

5-12-13 Pythagorean triple!

= sinθ + cosθ

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

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

= 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.

-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 - 2sinθcosθ = 1 - sin2θ

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

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

= sin θ

1-cos²θ = sin²θ.
√(sin²θ) = |sinθ| = sinθ (since 0°<θ<90°, sinθ>0).

గుర్తు: √(1-cos²θ) = sinθ, √(1-sin²θ) = cosθ — basic identity derivations!

tan 45° = 1

tan θ = sin θ / cos θ.
tan 45° = sin 45° / cos 45° = (1/√2) / (1/√2) = 1.

గుర్తు: tan values: 0, 1/√3, 1, √3, ∞ (0°,30°,45°,60°,90°).
tan 45° = 1 అంటే angle = 45° అయినప్పుడు opposite = adjacent!

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!

= 0

cos 90° = 0.
Any number × 0 = 0.
కాబట్టి మొత్తం product = 0.

Trick: cos 90° = 0 ఉన్నందువల్ల entire product zero!
ఇది shortcut — long calculation అక్కరలేదు.

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.

θ = 90°, 210°, 330°

Let s=sinθ: 2s²-s-1=0 → (2s+1)(s-1)=0.
s=1: sinθ=1 → θ=90°.
s=-1/2: sinθ=-1/2 → θ=210° or 330°.

sinθ=-1/2 in Q3 and Q4 (sin negative):
Q3: 180°+30°=210°. Q4: 360°-30°=330°.

= 1

1+tan²θ = sec²θ → 1/(1+tan²θ) = cos²θ.
f(θ) = sin²θ + cos²θ = 1.

గుర్తు: Trig expressions often simplify to constants using identities!

θ = 45°

sinθ = cosθ → sinθ/cosθ = 1 → tanθ = 1 → θ = 45°.

At θ=45°: sin45°=cos45°=1/√2.
This is the only angle in (0°,90°) where sin=cos!

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 ✓

cos²θ - sin²θ = cos 2θ

Double angle formula:
cos 2θ = cos²θ - sin²θ.
Also: cos 2θ = 2cos²θ - 1 = 1 - 2sin²θ.

ఉదా: cos²45°-sin²45°=1/2-1/2=0=cos 90° ✓.

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.