Trigonometry (త్రికోణమితి)

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!

= √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.

= 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] ✓.

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"!

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) ✓.

= 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).

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°!

-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!

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!

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° ✓.

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.

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!)

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.

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!

cos θ

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

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

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!

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=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) ✓.

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!

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!

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 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!

= 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.

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 = 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.

(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°).

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.

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!

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(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 ✓.

= 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 ✓.

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 ✓

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.

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).

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!

sin θ = 0

2sin²θ + 3cos²θ = 3.
2sin²θ + 3(1-sin²θ) = 3.
2sin²θ + 3 - 3sin²θ = 3.
-sin²θ = 0 → sin θ = 0.

θ = 0° లేదా 180°.

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.

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!

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 θ

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

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

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.

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

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.

= 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 ✓

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 ✓

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 = 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 = 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!

= 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)!

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!