Trigonometry (త్రికోణమితి)

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(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 = 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.

-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!

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.

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°).

= sinθ + cosθ

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

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

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°!

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!

= (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.

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 ✓

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!

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) ✓.

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!

= √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.

= 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)!

= 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 θ

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

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

= 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.

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). ✓

= 1/8

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.

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... ✓

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)).

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)) ✓.

(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°).

= 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 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.

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

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

గుర్తు: Trig expressions often simplify to constants using identities!

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 ✓.

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°.

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).

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...

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!

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!

= 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.

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!

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!

= 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).

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

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

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.

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.

= 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.

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!

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

(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)!

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!

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"!

= 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 ✓.

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.