🔄 Part 1 — Basic Trig Identities & Simplification
The Trig Identity Toolbox
Every identity you need for simplifying expressions
| Reciprocal Identities | Quotient Identities |
|---|---|
| sin x = 1/csc x · csc x = 1/sin x | tan x = sin x / cos x |
| cos x = 1/sec x · sec x = 1/cos x | cot x = cos x / sin x |
| tan x = 1/cot x · cot x = 1/tan x | — |
Strategy for simplifying trig expressions
Convert everything to sin and cos first. Then multiply/divide fractions, cancel common factors, and look for Pythagorean identity substitutions. Work methodically — one step at a time.
📌 Example 1A — f(x) = (sin x)(sec²x) / csc x. Rewrite using tan x only.
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Replace: sec²x = (1/cos x)², csc x = 1/sin x.2
f(x) = (sin x)(1/cos x)² ÷ (1/sin x) = (sin x)(1/cos²x) · (sin x) = sin²x / cos²x.3
sin x / cos x = tan x, so sin²x / cos²x = (sin x/cos x)² = tan²x.f(x) = tan²x
📌 Example 1B — g(x) = csc x · tan x. Rewrite using sec x only.
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Replace: csc x = 1/sin x, tan x = sin x/cos x.2
g(x) = (1/sin x)(sin x/cos x) = sin x / (sin x · cos x) = 1/cos x.3
1/cos x = sec x.g(x) = sec x
📌 Example 1C — f(x) = tan x / csc x. Rewrite as a fraction in powers of cos x only.
1
tan x = sin x/cos x, csc x = 1/sin x. Dividing: (sin x/cos x) ÷ (1/sin x) = (sin x/cos x)(sin x) = sin²x/cos x.2
Apply Pythagorean identity: sin²x = 1 − cos²x.f(x) = sin²x / cos x = (1 − cos²x) / cos x
📐 Part 2 — The Pythagorean Identities
Pythagorean Identities — Three Forms
From sin²+cos²=1, divide by cos² or sin² to get two more
Original (Unit Circle)
sin²θ + cos²θ = 1
x=cosθ, y=sinθ on unit circle radius 1. Pythagorean Theorem: x²+y²=1.
Divide by cos²θ
1 + tan²θ = sec²θ
sin²/cos² + cos²/cos² = 1/cos² → tan²θ + 1 = sec²θ
Divide by sin²θ
1 + cot²θ = csc²θ
sin²/sin² + cos²/sin² = 1/sin² → 1 + cot²θ = csc²θ
Useful rearrangements to memorize
From sin²+cos²=1: sin²x = 1−cos²x and cos²x = 1−sin²x
From 1+tan²=sec²: tan²x = sec²x−1
From 1+cot²=csc²: cot²x = csc²x−1
These rearrangements are very useful when substituting inside harder problems.
🔁 Part 3 — Inverse Trig Identities
arccos and arcsin — The Unit Circle Triangle
From x² + h² = 1, the missing side is √(1−x²)
📐 The Triangle Insight
On the unit circle: hypotenuse = 1, base = x = cos θ, height = h.
By Pythagorean Theorem: x² + h² = 1 → h = √(1−x²)
So sin θ = √(1−x²)/1 = √(1−x²), meaning θ = arcsin(√(1−x²)).
But also cos θ = x, meaning θ = arccos(x). So arccos(x) = arcsin(√(1−x²)).
By Pythagorean Theorem: x² + h² = 1 → h = √(1−x²)
So sin θ = √(1−x²)/1 = √(1−x²), meaning θ = arcsin(√(1−x²)).
But also cos θ = x, meaning θ = arccos(x). So arccos(x) = arcsin(√(1−x²)).
📋 Inverse Trig Identities
cos⁻¹(x) = sin⁻¹(√(1−x²))
arccos x = arcsin √(1−x²)
sin⁻¹(x) = cos⁻¹(√(1−x²))
arcsin x = arccos √(1−x²)
➕ Part 4 — Sum, Difference & Double Angle Identities
All the Formulas
The most tested identities on the AP exam
sin(α ± β)
sin(α+β) = sinα cosβ + cosα sinβ
sin(α−β) = sinα cosβ − cosα sinβ
Signs match: ± on left = ± in middle
cos(α ± β)
cos(α+β) = cosα cosβ − sinα sinβ
cos(α−β) = cosα cosβ + sinα sinβ
Signs FLIP: + on left → − in formula
Double Angle Identities
sin(2θ) = 2 sin θ cos θ
cos(2θ) = cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1
cos(2θ) has THREE equivalent forms — choose the one most useful for the problem
The crucial cos sign-flip rule
For sin, the sign in the middle matches what's outside: sin(α+β) has a + in the middle.
For cos, the sign flips: cos(α+β) has a − in the middle, and cos(α−β) has a + in the middle. This is the most commonly missed detail on exams.
📌 Example 2 — Which is equivalent to 2sin(π/14)cos(π/14)?
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Recognize the double angle pattern: 2sinθcosθ = sin(2θ).2
Here θ = π/14, so 2θ = 2·(π/14) = π/7.(A)
sin(π/28)
Would need θ = π/56 ✗
(B)
sin(π/7)cos(π/7)
That's only half the double angle result ✗
✓ (C)
sin(π/7)
2sin(π/14)cos(π/14) = sin(2·π/14) = sin(π/7) ✓
(D)
cos(π/7)
That's a cosine, not sine ✗
📌 Example 3 — k(x) = 4cos(2x). Which is equivalent?
1
Use cos(2x) = 2cos²x − 1 (the form that gives cos² only).2
k(x) = 4(2cos²x − 1) = 8cos²x − 4.(A)
8sinx cosx
That would be 4sin(2x), not 4cos(2x) ✗
(B)
4 − 8cos²x
Uses cos(2x)=1−2cos²x with wrong factor sign ✗
✓ (C)
8cos²x − 4
4(2cos²x−1) = 8cos²x−4 ✓
(D)
4cos²x + 4sin²x
= 4(cos²x+sin²x) = 4·1 = 4, not 4cos(2x) ✗
📌 Example 4 — Which is equivalent to cos(π/8)cos(π/16) − sin(π/8)sin(π/16)?
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Recognize the cos(α+β) pattern: cosαcosβ − sinαsinβ = cos(α+β).2
α = π/8, β = π/16. α + β = π/8 + π/16 = 2π/16 + π/16 = 3π/16.(A)
cos(π/16)
π/8 and π/16 add to 3π/16, not π/16 ✗
(B)
sin(π/16)
Wrong function — it's cosine addition ✗
✓ (C)
cos(3π/16)
cos(π/8+π/16) = cos(2π/16+π/16) = cos(3π/16) ✓
(D)
sin(3π/16)
Pattern matches cos addition, not sin ✗
⭕ Part 5 — Applying Identities with Circle Coordinates
Examples 5 & 6 — From Point to trig Values
Use (x, y) on a circle of radius r: sinθ = y/r, cosθ = x/r, then apply identities
Finding sin and cos from a point on a circle
If a terminal ray intersects a circle of radius r at point (x, y):
sin θ = y/r · cos θ = x/r
Find r using the Pythagorean theorem: r = √(x² + y²)
Then apply whatever double angle or sum/difference formula is needed.
📌 Example 5 — Circle at origin, P=(5,12). Find sin(2θ).
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r = √(5²+12²) = √(25+144) = √169 = 13.2
sinθ = 12/13, cosθ = 5/13.3
sin(2θ) = 2sinθcosθ = 2(12/13)(5/13) = 2·60/169 = 120/169.sin(2θ) = 120/169
📌 Example 6 — Two circles: P=(−5, √11) on radius-6 circle (angle α), Q=(2, √5) on radius-3 circle (angle β). Find cos(2α), sin(α+β), cos(α−β).
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Extract values:sinα = √11/6, cosα = −5/6 (P is in Q2 — negative x)
sinβ = √5/3, cosβ = 2/3
a) cos(2α)
cos(2α) = 2cos²α − 1
= 2(−5/6)² − 1 = 2(25/36) − 1 = 50/36 − 36/36 = 14/36 = 7/18
= 2(−5/6)² − 1 = 2(25/36) − 1 = 50/36 − 36/36 = 14/36 = 7/18
b) sin(α + β)
sin(α+β) = sinα cosβ + cosα sinβ
= (√11/6)(2/3) + (−5/6)(√5/3)
= 2√11/18 − 5√5/18 = (2√11 − 5√5)/18
= (√11/6)(2/3) + (−5/6)(√5/3)
= 2√11/18 − 5√5/18 = (2√11 − 5√5)/18
c) cos(α − β)
cos(α−β) = cosα cosβ + sinα sinβ
= (−5/6)(2/3) + (√11/6)(√5/3)
= −10/18 + √55/18 = (√55 − 10)/18
= (−5/6)(2/3) + (√11/6)(√5/3)
= −10/18 + √55/18 = (√55 − 10)/18
Quick Reference — All Identities
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📐 Pythagorean Identities
sin²x + cos²x = 1
1 + tan²x = sec²x (÷ by cos²)
1 + cot²x = csc²x (÷ by sin²)
sin²x = 1−cos²x · cos²x = 1−sin²x
➕ Sum/Difference & Double Angle
sin(α±β) = sinαcosβ ± cosαsinβ
cos(α±β) = cosαcosβ ∓ sinαsinβ
sin(2θ) = 2sinθcosθ
cos(2θ) = cos²θ−sin²θ = 1−2sin²θ = 2cos²θ−1
⚠️ Common Mistakes
❌ sin(α+β) = sinα + sinβ
Wrong! sin(α+β) = sinαcosβ + cosαsinβ. You can't just distribute into sin.
❌ cos(2θ) = 2cosθ
Wrong! cos(2θ) = 2cos²θ−1. The double angle changes the expression significantly.
❌ cos(α+β) has + in the middle
cos FLIPS: cos(α+β) = cosαcosβ − sinαsinβ. The sign flips to minus.
❌ sin²x + cos²x = 2 or 0
Always equals exactly 1 for any angle. This is one of the most fundamental trig facts.