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3.12 · Trig Reciprocal Identities
State the three reciprocal identity pairs: sin/csc, cos/sec, tan/cot.
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3.12 · Quotient Identities
Write tan x and cot x each as a fraction using sin x and cos x.
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3.12 · Simplify: csc·tan
Simplify: csc x · tan x. Write using only sec x.
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3.12 · Simplify: sin·sec²/csc
Simplify: (sin x)(sec²x)/csc x. Write using only tan x.
sin²x = 1−cos²x
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3.12 · Simplify: tan/csc
Simplify: tan x / csc x. Write as a fraction using sin and cos only.
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3.12 · Pythagorean Identity 1
State the fundamental Pythagorean identity. Where does it come from?
Unit circle: x=cosθ, y=sinθ, r=1
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3.12 · Pythagorean Identity 2
Derive 1 + tan²θ = sec²θ from sin²θ + cos²θ = 1.
Divide every term by cos²θ
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3.12 · Pythagorean Identity 3
Derive 1 + cot²θ = csc²θ from sin²θ + cos²θ = 1.
Divide every term by sin²θ
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✓ Quotient Identities
tan x = sin x / cos x · cot x = cos x / sin x
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✓ Trig Reciprocal Identities
sin = 1/csc, csc = 1/sin · cos = 1/sec, sec = 1/cos · tan = 1/cot, cot = 1/tan
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✓ Simplify: sin·sec²/csc
(sin x)(1/cos²x)/(1/sin x) = sin²x/cos²x = tan²x
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✓ Simplify: csc·tan
csc x · tan x = (1/sin x)(sin x/cos x) = 1/cos x = sec x
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✓ Pythagorean Identity 1
sin²θ + cos²θ = 1. Comes from the unit circle: x² + y² = r², with x=cosθ, y=sinθ, r=1.
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✓ Simplify: tan/csc
(sin x/cos x)/(1/sin x) = sin²x/cos x = (1−cos²x)/cos x
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✓ Pythagorean Identity 3
Divide sin²θ+cos²θ=1 by sin²θ: 1 + cot²θ = csc²θ
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✓ Pythagorean Identity 2
Divide sin²θ+cos²θ=1 by cos²θ: tan²θ + 1 = sec²θ, or 1 + tan²θ = sec²θ
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3.12 · Pythagorean Rearrangements
Write 4 useful rearrangements: sin²x=?, cos²x=?, tan²x=?, sec²x=?
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3.12 · arccos + arcsin Identity
State the identity connecting arccos(x) and arcsin(x).
They add up to...
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3.12 · Cofunction Identity
State the cofunction identity: sin(π/2 − x) = ? and cos(π/2 − x) = ?
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3.12 · cos Sum/Difference Formula
State cos(α+β) and cos(α−β).
Signs are opposite to what you'd expect
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3.12 · sin Double Angle
State sin(2θ) in terms of sin and cos.
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3.12 · cos Double Angle — 3 Forms
State all THREE forms of cos(2θ).
Use Pythagorean identity to derive forms 2 and 3
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3.12 · sin Sum/Difference Formula
State sin(α+β) and sin(α−β).
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3.12 · Apply: 4cos(2x)
Which is equivalent to 4cos(2x)? A) 4−8sin²x B) 8sin²x−4 C) 8cos²x−4 D) 4−8cos²x
Use cos(2θ) = 2cos²θ−1
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✓ arccos + arcsin Identity
arccos(x) + arcsin(x) = π/2. They are complementary — the angles sum to 90°.
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✓ Pythagorean Rearrangements
sin²x = 1−cos²x · cos²x = 1−sin²x · tan²x = sec²x−1 · sec²x = 1+tan²x
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✓ cos Sum/Difference Formula
cos(α+β) = cosαcosβ − sinαsinβ · cos(α−β) = cosαcosβ + sinαsinβ
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✓ Cofunction Identity
sin(π/2 − x) = cos x · cos(π/2 − x) = sin x. Cofunctions of complementary angles are equal.
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✓ cos Double Angle — 3 Forms
cos(2θ) = cos²θ−sin²θ = 1−2sin²θ = 2cos²θ−1
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✓ sin Double Angle
sin(2θ) = 2sinθcosθ
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✓ Apply: 4cos(2x)
4cos(2x) = 4(2cos²x−1) = 8cos²x−4 → Answer: (C) 8cos²x−4
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✓ sin Sum/Difference Formula
sin(α+β) = sinαcosβ + cosαsinβ · sin(α−β) = sinαcosβ − cosαsinβ
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3.12 · Point on Circle — sin, cos
P = (5, 12) on its circle (r=13). Find sinθ, cosθ.
r = √(x²+y²)
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3.12 · Point on Circle — sin(2θ)
P = (5, 12) on its circle (r=13). Find sin(2θ).
sin(2θ) = 2sinθcosθ
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3.12 · Verify Trig Identity
Verify: (sin x)(sec x) = tan x.
Start from the more complex side
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3.12 · Sum Formula from Two Points
P=(−5,√11) on r=6 circle (angle α), Q=(2,√5) on r=3 circle (angle β). Find sin(α+β).
sin(α+β) = sinαcosβ + cosαsinβ
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✓ Point on Circle — sin(2θ)
sin(2θ) = 2sinθcosθ = 2(12/13)(5/13) = 120/169
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✓ Point on Circle — sin, cos
r = √(25+144) = 13. sinθ = 12/13, cosθ = 5/13
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✓ Sum Formula from Two Points
sinα=√11/6, cosα=−5/6, sinβ=√5/3, cosβ=2/3 · sin(α+β) = (√11/6)(2/3)+(−5/6)(√5/3) = (2√11−5√5)/18
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✓ Verify Trig Identity
(sin x)(sec x) = (sin x)(1/cos x) = sin x/cos x = tan x ✓
Start from one side only, transform to match the other.