Every identity, formula, and strategy you need for Chapter 7 — from fundamental reciprocal identities to inverse trig functions and solving equations on any interval.
These eight identities are the foundation of every identity proof in Chapter 7. Memorize them cold — everything else is built from these.
| Function | Reciprocal | Memory Tip |
|---|---|---|
| sin θ | csc θ = 1 / sin θ | co-secant flips sine |
| cos θ | sec θ = 1 / cos θ | secant flips cosine |
| tan θ | cot θ = 1 / tan θ | cotangent flips tangent |
tan θ = sin θ / cos θ
Valid whenever cos θ ≠ 0
cot θ = cos θ / sin θ
Valid whenever sin θ ≠ 0
All three come from the unit circle equation x² + y² = 1. Learn to derive forms 2 and 3 from form 1 by dividing through by cos²θ or sin²θ.
sin²θ + cos²θ = 1
Unit circle x² + y² = 1
sin²θ = 1 − cos²θ
cos²θ = 1 − sin²θ
1 + tan²θ = sec²θ
Divide row 1 by cos²θ
tan²θ = sec²θ − 1
sec²θ − tan²θ = 1
1 + cot²θ = csc²θ
Divide row 1 by sin²θ
cot²θ = csc²θ − 1
csc²θ − cot²θ = 1
Exam Trick
On any test that asks for a Pythagorean identity, if you can only remember sin²θ + cos²θ = 1, you can always reconstruct the other two in 10 seconds by dividing. Never try to memorize 1 + tan²θ = sec²θ as a separate fact — derive it on the spot.
Verification is different from algebra: you must show the two sides are equal without treating the equation as something you can solve. Work ONE side only.
Strategies that work
Common mistakes to avoid
Worked Example 1
Verify: (sin θ / (1 − cos θ)) = (1 + cos θ) / sin θ
Work the left side (LHS).
LHS = sin θ / (1 − cos θ)
// Multiply by conjugate (1 + cos θ)/(1 + cos θ)
= sin θ (1 + cos θ) / [(1 − cos θ)(1 + cos θ)]
// Denominator is difference of squares: 1 − cos²θ
= sin θ (1 + cos θ) / (1 − cos²θ)
// Pythagorean: 1 − cos²θ = sin²θ
= sin θ (1 + cos θ) / sin²θ
// Cancel one sin θ
= (1 + cos θ) / sin θ = RHS ✓
Worked Example 2 — Converting to sin/cos
Verify: tan²θ − sin²θ = tan²θ · sin²θ
LHS = tan²θ − sin²θ
// Replace tan²θ = sin²θ/cos²θ
= sin²θ/cos²θ − sin²θ
// Factor out sin²θ
= sin²θ (1/cos²θ − 1)
// 1/cos²θ − 1 = (1 − cos²θ)/cos²θ = sin²θ/cos²θ
= sin²θ · (sin²θ/cos²θ)
= sin²θ · tan²θ = RHS ✓
Simplification means rewriting an expression in a cleaner or specified form. Unlike verification, there is no RHS to match — you stop when the expression is as simple as the problem intends (often a single trig function or a constant).
Example A: Simplify to a single trig function
sin θ · cot θ · sec θ
// Replace cot = cos/sin, sec = 1/cos
= sin θ · (cos θ / sin θ) · (1 / cos θ)
// Cancel sin θ and cos θ
= 1
Example B: Simplify using Pythagorean identity
cos²θ (1 + tan²θ)
// 1 + tan²θ = sec²θ
= cos²θ · sec²θ
// sec²θ = 1/cos²θ
= cos²θ / cos²θ = 1
Example C: Simplify by factoring
(sin x + cos x)² − 1
// Expand: sin²x + 2 sin x cos x + cos²x − 1
= (sin²x + cos²x) + 2 sin x cos x − 1
// sin²x + cos²x = 1
= 1 + 2 sin x cos x − 1
= 2 sin x cos x = sin(2x)
These formulas let you find exact values for angles like 15°, 75°, and 105° by splitting them into angles whose trig values you already know (30°, 45°, 60°, 90°).
Sine
sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
Signs STAY the same as the ± outside
Cosine
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
Signs FLIP — cosine is contrary
Tangent
tan(A + B) = (tan A + tan B) / (1 − tan A tan B)
tan(A − B) = (tan A − tan B) / (1 + tan A tan B)
Denominator sign also flips
Worked Example — Exact value of sin(75°)
sin(75°) = sin(45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
= (√2/2)(√3/2) + (√2/2)(1/2)
= √6/4 + √2/4
= (√6 + √2) / 4
Double angle formulas follow from the sum formulas by setting A = B. The cosine double angle formula has three useful forms — know all three and when each is most convenient.
sin(2A)
2 sin A cos A
Only one form — no choices needed
cos(2A) — three forms
cos²A − sin²A
2 cos²A − 1
1 − 2 sin²A
tan(2A)
2 tan A / (1 − tan²A)
Undefined when tan A = ±1
Power-Reducing Forms
The second and third forms of cos(2A) rearrange to give power-reducing identities, which appear heavily in calculus integration:
sin²A = (1 − cos 2A) / 2
cos²A = (1 + cos 2A) / 2
Worked Example — Find sin(2θ) given sin θ = 3/5, θ in Q2
// In Q2, cos θ < 0. Use sin²θ + cos²θ = 1
cos²θ = 1 − (3/5)² = 1 − 9/25 = 16/25
cos θ = −4/5 (negative in Q2)
// Apply double angle formula
sin(2θ) = 2 sin θ cos θ = 2(3/5)(−4/5) = −24/25
Half angle formulas let you find exact values for angles like 22.5°, 112.5°, and π/8. The ± sign is determined by the quadrant of A/2, not the quadrant of A.
sin(A/2)
= ±√[(1 − cos A) / 2]
+ if A/2 is in Q1 or Q2 (sin positive)
cos(A/2)
= ±√[(1 + cos A) / 2]
+ if A/2 is in Q1 or Q4 (cos positive)
tan(A/2)
= ±√[(1 − cos A) / (1 + cos A)]
= sin A / (1 + cos A)
= (1 − cos A) / sin A
Last two forms have no ± ambiguity
Worked Example — Exact value of sin(22.5°)
// 22.5° = 45°/2, so A = 45°
sin(22.5°) = ±√[(1 − cos 45°) / 2]
// 22.5° is in Q1, so sin is positive. cos 45° = √2/2
= √[(1 − √2/2) / 2]
= √[(2 − √2) / 4]
= √(2 − √2) / 2
These formulas are less commonly memorized but appear on Chapter 7 tests and become critical in calculus integration. Know where they come from (adding and subtracting the sum/difference formulas) so you can reconstruct them if needed.
sin A cos B = (1/2)[sin(A + B) + sin(A − B)]
cos A sin B = (1/2)[sin(A + B) − sin(A − B)]
cos A cos B = (1/2)[cos(A + B) + cos(A − B)]
sin A sin B = (1/2)[cos(A − B) − cos(A + B)]
Memory: these come from ADDING or SUBTRACTING the sum and difference formulas.
sin A + sin B = 2 sin[(A+B)/2] cos[(A−B)/2]
sin A − sin B = 2 cos[(A+B)/2] sin[(A−B)/2]
cos A + cos B = 2 cos[(A+B)/2] cos[(A−B)/2]
cos A − cos B = −2 sin[(A+B)/2] sin[(A−B)/2]
Use sum-to-product to factor a trig equation that looks like sin x + sin(3x) = 0.
Inverse trig functions answer the question: which angle has this trig value? Because trig functions are not one-to-one on their natural domains, we restrict the domain to get a unique output — called the principal value.
| Function | Notation | Domain | Range (Principal Values) |
|---|---|---|---|
| Inverse Sine | arcsin x or sin⁻¹ x | [−1, 1] | [−π/2, π/2] |
| Inverse Cosine | arccos x or cos⁻¹ x | [−1, 1] | [0, π] |
| Inverse Tangent | arctan x or tan⁻¹ x | (−∞, ∞) | (−π/2, π/2) |
| Inverse Cosecant | arccsc x | (−∞,−1] ∪ [1, ∞) | [−π/2, 0) ∪ (0, π/2] |
| Inverse Secant | arcsec x | (−∞,−1] ∪ [1, ∞) | [0, π/2) ∪ (π/2, π] |
| Inverse Cotangent | arccot x | (−∞, ∞) | (0, π) |
Cancellation — when it works
sin(arcsin x) = x, for x in [−1, 1]
arcsin(sin x) = x, ONLY for x in [−π/2, π/2]
cos(arccos x) = x, for x in [−1, 1]
arccos(cos x) = x, ONLY for x in [0, π]
tan(arctan x) = x, for all x
arctan(tan x) = x, ONLY for x in (−π/2, π/2)
The classic exam trap
arcsin(sin(5π/6)) is NOT 5π/6.
Because 5π/6 is not in [−π/2, π/2], the cancellation law does not apply. Instead: sin(5π/6) = 1/2, then arcsin(1/2) = π/6.
Answer: π/6
Worked Example — Evaluating a composition: cos(arctan(3/4))
// Let θ = arctan(3/4), so tan θ = 3/4 with θ in (−π/2, π/2)
// Draw a right triangle: opposite = 3, adjacent = 4
hypotenuse = √(3² + 4²) = √25 = 5
cos θ = adjacent/hypotenuse = 4/5
Therefore cos(arctan(3/4)) = 4/5
A trig equation is true for specific values of the variable. You need two things: all solutions on [0, 2π) for restricted-domain problems, and the general solution (with period multiples) for open-domain problems.
5-Step Process for Solving Any Trig Equation
Which quadrants are positive?
Q1
All positive
Q2
sin positive
Q3
tan positive
Q4
cos positive
Mnemonic: All Students Take Calculus (Q1 Q2 Q3 Q4)
How many solutions on [0, 2π)?
sin x = c: 2 solutions if |c| < 1, one in Q1/Q2 for c > 0, one in Q3/Q4 for c < 0
cos x = c: 2 solutions if |c| < 1, one in Q1/Q4 for c > 0, one in Q2/Q3 for c < 0
tan x = c: exactly 1 solution per period (π), always 2 on [0, 2π)
Example A — Solutions on [0, 2π): 2 sin x − √3 = 0
2 sin x = √3
sin x = √3/2
// sin is positive, so Q1 and Q2. Reference angle: arcsin(√3/2) = π/3
x = π/3 (Q1) or x = π − π/3 = 2π/3 (Q2)
Solution set: x = π/3, 2π/3
Example B — Quadratic: 2 sin²x − sin x − 1 = 0 on [0, 2π)
// Let u = sin x
2u² − u − 1 = 0
(2u + 1)(u − 1) = 0
u = −1/2 or u = 1
// sin x = −1/2: Q3 and Q4
x = π + π/6 = 7π/6 or x = 2π − π/6 = 11π/6
// sin x = 1
x = π/2
Solution set: x = π/2, 7π/6, 11π/6
Example C — Using an identity: cos(2x) = sin x on [0, 2π)
// Use cos(2x) = 1 − 2sin²x to get everything in sin
1 − 2 sin²x = sin x
2 sin²x + sin x − 1 = 0
(2 sin x − 1)(sin x + 1) = 0
sin x = 1/2 → x = π/6, 5π/6
sin x = −1 → x = 3π/2
Solution set: x = π/6, 5π/6, 3π/2
Example D — General solution: tan x = −1
// Reference angle: arctan(1) = π/4. tan is negative in Q2 and Q4.
Solutions in [0, π): x = 3π/4 (Q2)
// Tan has period π, so general solution:
x = 3π/4 + πk, where k is any integer
| Category | Formula(s) |
|---|---|
| Pythagorean #1 | sin²θ + cos²θ = 1 |
| Pythagorean #2 | 1 + tan²θ = sec²θ |
| Pythagorean #3 | 1 + cot²θ = csc²θ |
| sin(A ± B) | sin A cos B ± cos A sin B |
| cos(A ± B) | cos A cos B ∓ sin A sin B |
| tan(A ± B) | (tan A ± tan B) / (1 ∓ tan A tan B) |
| sin(2A) | 2 sin A cos A |
| cos(2A) form 1 | cos²A − sin²A |
| cos(2A) form 2 | 2cos²A − 1 |
| cos(2A) form 3 | 1 − 2sin²A |
| tan(2A) | 2 tan A / (1 − tan²A) |
| sin(A/2) | ±√[(1 − cos A) / 2] |
| cos(A/2) | ±√[(1 + cos A) / 2] |
| tan(A/2) | (1 − cos A) / sin A = sin A / (1 + cos A) |
| sin²A (power) | (1 − cos 2A) / 2 |
| cos²A (power) | (1 + cos 2A) / 2 |
| sin A cos B | (1/2)[sin(A+B) + sin(A−B)] |
| sin A sin B | (1/2)[cos(A−B) − cos(A+B)] |
| cos A cos B | (1/2)[cos(A+B) + cos(A−B)] |
| sin A + sin B | 2 sin[(A+B)/2] cos[(A−B)/2] |
| cos A + cos B | 2 cos[(A+B)/2] cos[(A−B)/2] |
Precalculus Study Guide
Complete roadmap for all Stewart Precalculus chapters with topic summaries and practice problem links.
Right Triangle Trig
SOH-CAH-TOA, special triangles, and exact values — the foundation you need before identities.
The Unit Circle
All 16 standard angles with exact sin, cos, and tan values — essential for solving trig equations by inspection.
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