Trigonometry (త్రికోణమితి)

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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!

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... ✓

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 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 ✓

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.

sec²θ - tan²θ = 1

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

Similarly: cosec²θ - cot²θ = 1.

= 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.

= 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.

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!

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+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)) ✓.

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!)

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.

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). ✓

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!

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...

= 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] ✓.

= √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.

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) ✓.

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!

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°!

= 1/16

cos 108° = cos(180°-72°) = -cos72°.
cos 144° = cos(180°-36°) = -cos36°.
Product = cos36° × cos72° × (-cos72°) × (-cos36°).
= cos²36° × cos²72°.
cos36°=(1+√5)/4×2...
Actually: product = 1/16.

These are roots of Chebyshev polynomial. Standard result: 1/16.

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°.

= 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 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.

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!

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).

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!

= 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.

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.

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°.

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!

= 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.

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 ✓

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!

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!

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°).

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!

= (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.

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.

(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

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.

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).

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 + 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

(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)!

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) ✓.

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.