Trigonometry (త్రికోణమితి)

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!

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.

Q1:+++ Q2:+--, Q3:-+-, Q4:--+... correctly

ASTC:
Q1 (0°-90°): sin+, cos+, tan+. (All positive)
Q2 (90°-180°): sin+, cos-, tan-. (Sin positive)
Q3 (180°-270°): sin-, cos-, tan+. (Tan positive)
Q4 (270°-360°): sin-, cos+, tan-. (Cos positive)

Remember: "All Students Take Calculus" or "Add Sugar To Coffee"!

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!

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!

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!)

sin30° = √((1-cos60°)/2) = √((1-½)/2) = √(1/4) = 1/2 ✓

sin(60°/2) = sin30° = 1/2.
RHS = √((1-cos60°)/2) = √((1-1/2)/2) = √(1/4) = 1/2 ✓.

Half angle formulas:
sin(θ/2) = ±√((1-cosθ)/2).
cos(θ/2) = ±√((1+cosθ)/2).
tan(θ/2) = sinθ/(1+cosθ) = (1-cosθ)/sinθ.

= 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.

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).

= 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.

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!

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!

0 + 0 = 0

sin 0° = 0.
cos 90° = 0.
Sum = 0 + 0 = 0.

గుర్తు: sin 0°=0 (no height at 0°), cos 90°=0 (no horizontal at 90°).
These are boundary values — very commonly tested!

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!

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!

= 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).

= 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 ✓.

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 ✓

= 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 ✓

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).

cos θ

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

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

= 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 ✓

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° ✓.

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!)

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!

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!

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 ✓.

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°!

= (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.

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.

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°).

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+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°).

= 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.

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

(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)!

= 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

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!

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.

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!

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!

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 90° = 0 ✓

cos(45°+45°) = cos45°cos45° - sin45°sin45°.
= (1/√2)(1/√2) - (1/√2)(1/√2).
= 1/2 - 1/2 = 0 ✓.

cos 90° = 0 confirmed by addition formula!

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°!

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°.

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 ✓

= 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!

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!

= 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] ✓.