1/20
3.12 · Reciprocal Identities
Write all 6 reciprocal identity pairs (sin↔csc, cos↔sec, tan↔cot).
Each function × its reciprocal = 1
2/20
3.12 · Quotient Identities
Write tan x and cot x each as a fraction using sin and cos only.
3/20
3.12 · Simplify: csc·tan
Simplify: csc x · tan x. Write using only sec x.
Replace each with sin/cos
4/20
3.12 · Simplify: sin·sec²/csc
Simplify: (sin x)(sec²x)/csc x. Write using only tan x.
5/20
3.12 · Simplify: tan/csc
Simplify: tan x / csc x. Write as a fraction using cos x only (apply Pythagorean identity).
sin²x = 1−cos²x
6/20
3.12 · Pythagorean Identity 1
State the fundamental Pythagorean identity. Where does it come from?
Unit circle: x=cosθ, y=sinθ, radius=1
7/20
3.12 · Pythagorean Identity 2
Derive 1+tan²θ=sec²θ from sin²θ+cos²θ=1.
Divide everything by cos²θ
8/20
3.12 · Pythagorean Identity 3
Derive 1+cot²θ=csc²θ from sin²θ+cos²θ=1.
Divide everything by sin²θ
2/20
✓ Quotient Identities
tan x = sin x / cos x · cot x = cos x / sin x
These follow directly from the unit circle: tan θ = y/x = sinθ/cosθ, and cot is its reciprocal.
1/20
✓ Reciprocal Identities
sin=1/csc · csc=1/sin · cos=1/sec · sec=1/cos · tan=1/cot · cot=1/tan
Each pair multiplies to 1. Memory: co-secant pairs with sine (non-co); secant pairs with cosine.
4/20
✓ Simplify sin·sec²/csc
= tan²x
(sinx)(1/cos²x) ÷ (1/sinx) = (sinx)(1/cos²x)(sinx) = sin²x/cos²x = (sinx/cosx)² = tan²x
3/20
✓ Simplify csc·tan
= sec x
(1/sinx)(sinx/cosx) = 1/cosx = sec x. The sin x factors cancel.
6/20
✓ Pythagorean Identity 1
sin²θ + cos²θ = 1
From unit circle: point (cosθ, sinθ) lies on circle of radius 1. Pythagorean Theorem: x²+y²=r² → cos²θ+sin²θ=1.
5/20
✓ Simplify tan/csc
= (1−cos²x)/cos x
tanx/cscx = (sinx/cosx)÷(1/sinx) = sin²x/cosx. Then use sin²x=1−cos²x.
8/20
✓ Pythagorean Identity 3
1 + cot²θ = csc²θ
Divide sin²θ+cos²θ=1 by sin²θ: 1 + cos²θ/sin²θ = 1/sin²θ → 1+cot²θ=csc²θ. Rearranged: cot²θ=csc²θ−1.
7/20
✓ Pythagorean Identity 2
1 + tan²θ = sec²θ
Divide sin²θ+cos²θ=1 by cos²θ: sin²θ/cos²θ + 1 = 1/cos²θ → tan²θ+1=sec²θ. Rearranged: tan²θ=sec²θ−1.
9/20
3.12 · Pyth Rearrangements
Write 4 useful rearrangements: sin²x=?, cos²x=?, tan²x=?, cot²x=?
Isolate each from the 3 Pythagorean identities
10/20
3.12 · Inverse Trig Identity
State the identity connecting arccos(x) and arcsin(x).
Think: unit circle right triangle with base x, hypotenuse 1
11/20
3.12 · sin Sum Formula
State sin(α+β) and sin(α−β).
Signs match in sin: + gives +, − gives −
12/20
3.12 · cos Sum Formula
State cos(α+β) and cos(α−β).
cos FLIPS the sign!
13/20
3.12 · Double Angle: sin
State sin(2θ) in terms of sin and cos.
Comes from sin(θ+θ)
14/20
3.12 · Double Angle: cos — 3 forms
State all THREE forms of cos(2θ).
One has only sin, one only cos, one has both
15/20
3.12 · Apply: 2sinθcosθ
Simplify: 2sin(π/14)cos(π/14).
Recognize double angle sin(2θ)=2sinθcosθ
16/20
3.12 · Apply: 4cos(2x)
Which is equivalent to 4cos(2x)? Choose: 8sinxcosx · 8cos²x−4 · 4−8cos²x
cos(2x) = 2cos²x − 1
10/20
✓ Inverse Trig Identity
arccos(x) = arcsin(√(1−x²)) and arcsin(x) = arccos(√(1−x²))
From unit circle right triangle: hypotenuse=1, base=x=cosθ, height=√(1−x²)=sinθ. So θ=arccos(x)=arcsin(√(1−x²)).
9/20
✓ Pythagorean Rearrangements
sin²x=1−cos²x · cos²x=1−sin²x · tan²x=sec²x−1 · cot²x=csc²x−1
These are essential substitution tools. When you see sin² in a cos problem, replace it with 1−cos².
12/20
✓ cos Sum Formula
cos(α+β) = cosαcosβ − sinαsinβ (sign FLIPS to −) · cos(α−β) = cosαcosβ + sinαsinβ (sign FLIPS to +)
The sign in cos ALWAYS flips: + outside → − inside, − outside → + inside. This is opposite to sin.
11/20
✓ sin Sum Formula
sin(α+β) = sinαcosβ + cosαsinβ · sin(α−β) = sinαcosβ − cosαsinβ
For sin: the sign in the MIDDLE matches the ± on the left. Easy to remember — no flipping.
14/20
✓ Double Angle: cos — 3 Forms
cos(2θ) = cos²θ−sin²θ = 1−2sin²θ = 2cos²θ−1
Derived by applying cos(α+β) with α=β=θ. The three forms come from substituting cos²=1−sin² or sin²=1−cos².
13/20
✓ Double Angle: sin
sin(2θ) = 2sinθcosθ
Derived from sin(α+β) with α=β=θ: sinθcosθ+cosθsinθ = 2sinθcosθ. Recognize this pattern in reverse too: 2sinxcosx = sin(2x).
16/20
✓ Apply: 4cos(2x)
8cos²x − 4 (answer C)
cos(2x)=2cos²x−1 → 4cos(2x)=4(2cos²x−1)=8cos²x−4. Not 8sinxcosx (that's 4sin(2x)), not 4−8cos²x.
15/20
✓ Apply: 2sin(π/14)cos(π/14)
= sin(π/7) (answer C)
Recognize 2sinθcosθ=sin(2θ). Here θ=π/14, so 2θ=π/7.
17/20
3.12 · Apply: cos·cos − sin·sin
Simplify: cos(π/8)cos(π/16) − sin(π/8)sin(π/16).
Recognize cos(α+β) = cosαcosβ − sinαsinβ
18/20
3.12 · Point on Circle — sin/cos
P=(5,12) is on a circle centered at origin. Find sinθ and cosθ.
Find radius r first: r=√(x²+y²)
19/20
3.12 · Double Angle from Point
P=(5,12) on its circle. Find sin(2θ).
Use sin(2θ)=2sinθcosθ
20/20
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).
Step 1: find sinα,cosα,sinβ,cosβ. Step 2: apply formula.
18/20
✓ Point on Circle — sinθ, cosθ
sinθ=12/13 · cosθ=5/13
r=√(5²+12²)=√169=13. sinθ=y/r=12/13. cosθ=x/r=5/13. (5,12,13 is a Pythagorean triple!)
17/20
✓ Apply: cos(π/8)cos(π/16)−sin(π/8)sin(π/16)
= cos(3π/16) (answer C)
Pattern cosαcosβ−sinαsinβ=cos(α+β). α=π/8=2π/16, β=π/16. α+β=3π/16.
20/20
✓ Sum Formula from Two Points
sin(α+β) = (2√11−5√5)/18
sinα=√11/6, cosα=−5/6, sinβ=√5/3, cosβ=2/3. sin(α+β)=(√11/6)(2/3)+(−5/6)(√5/3)=2√11/18−5√5/18.
19/20
✓ Double Angle from Point
sin(2θ) = 120/169
sinθ=12/13, cosθ=5/13. sin(2θ)=2sinθcosθ=2(12/13)(5/13)=120/169.