Math Base — High-School Reference

An interactive companion to the Math/base notes. Pick a subject folder, then a topic.

2.1Identities

Pythagorean

  • sin²θ + cos²θ = 1
  • 1 + tan²θ = sec²θ
  • 1 + cot²θ = csc²θ

Quotient & reciprocal

  • tan θ = sin θ / cos θ
  • cot θ = cos θ / sin θ = 1/tan θ
  • csc θ = 1/sin θ,  sec θ = 1/cos θ

Sum / difference

  • 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)

Double angle

  • sin 2θ = 2 sin θ cos θ
  • cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ
  • tan 2θ = 2 tan θ / (1 − tan²θ)

Half angle

  • sin²(θ/2) = (1 − cos θ) / 2
  • cos²(θ/2) = (1 + cos θ) / 2

Co-function & even/odd

  • sin(90° − θ) = cos θ,  cos(90° − θ) = sin θ
  • sin(−θ) = −sin θ (odd)
  • cos(−θ) = cos θ (even),  tan(−θ) = −tan θ (odd)

Identity checker

Pick an identity and an angle — both sides are evaluated numerically so you can see they always match.

left-hand side– right-hand side– match?–

2.2The unit circle

A circle of radius 1 centred at the origin. For an angle θ measured counter-clockwise from the positive x-axis, the terminal side meets the circle at (cos θ, sin θ). So cos θ is the x-coordinate, sin θ is the y-coordinate, and tan θ = sin θ / cos θ is undefined wherever cos θ = 0.

Quadrant signs ("All Students Take Calculus", I → II → III → IV): QI all positive; QII only sin; QIII only tan; QIV only cos.

Where tan θ sits: draw the vertical tangent line at (1, 0); extend the radius for θ until it crosses that line — the segment from (1,0) to the crossing is exactly tan θ. That's where the name comes from.

Drag the circle

I II III IV x = 1 (1,0)
θ– sin θ– cos θ– tan θ– quadrant– signs (sin,cos,tan)–

Drag the purple point around the circle, or the green point along the tangent line.

Exact values

Click a column to place that angle on the circle above.

θ0°30°45°60°90°
sin θ01/2√2/2√3/21
cos θ1√3/2√2/21/20
tan θ0√3/31√3undef.

2.3Radians

A radian is the angle whose arc length equals the radius (arc = r). A full circle is 360° = 2π rad. Degrees → radians: × π/180. Radians → degrees: × 180/π.

Degree / radian converter

degrees60° radians (decimal)1.047 radians (× π)π/3 arc length (r = 1)1.047
Degrees0°30°45°60°90°180°270°360°
Radians0π/6π/4π/3π/2π3π/22π

2.4Inverse trig — arcsin, arccos, arctan

sin, cos, tan are not one-to-one over all reals, so each is first restricted to a monotonic interval. The inverses return a single principal value, not every angle with that ratio.

FunctionDomainRange (principal values)
arcsin(k)k ∈ [−1, 1][−π/2, π/2] — right half of the circle
arccos(k)k ∈ [−1, 1][0, π] — top half of the circle
arctan(k)k ∈ ℝ(−π/2, π/2) — open, since tan(±π/2) is undefined

Principal-value explorer

defined for this k?yes principal value– in degrees–

Because each arc-function hands back only one representative angle, they don't by themselves solve sin x = k — that's why the general solutions below add +2πn and a second mirrored branch.

2.5Periodicity

sin and cos repeat every 2π (360°): sin(θ + 2πn) = sin θ. tan repeats every π (180°): tan(θ + πn) = tan θ.

See the repeat

f(x)– f(x + period)– period used2π

2.6Solving trig equations

General solutions (n is any integer):

  • sin x = k → x = arcsin k + 2πn  or  x = π − arcsin k + 2πn
  • cos x = k → x = ± arccos k + 2πn
  • tan x = k → x = arctan k + πn  (one family — tan has period π)

General-solution generator

Use a compound argument a·x + b   

e.g. for sin(2x + π/3) = ½ set a = 2, b ≈ 1.0472.

Trivial zero cases need only one family each: sin x = 0 → x = πn;  cos x = 0 → x = π/2 + πn;  tan x = 0 → x = πn. At x = πn: sin = 0 and cos = (−1)ⁿ. At x = π/2 + πn: it flips — cos = 0 and sin = (−1)ⁿ.
Worked example — sin(2x + π/3) = 1/2

Let u = 2x + π/3. Then sin u = 1/2 → u = π/6 + 2πn or u = 5π/6 + 2πn.

Branch 1: 2x + π/3 = π/6 + 2πn → 2x = −π/6 + 2πn → x = −π/12 + πn.

Branch 2: 2x + π/3 = 5π/6 + 2πn → 2x = π/2 + 2πn → x = π/4 + πn.

Answer: x = −π/12 + πn or x = π/4 + πn. The period in x shrank to π = 2π/2 because of the coefficient 2.

AAbsolute value

Solving equations with |·|method + solver
  1. Isolate the absolute value so it stands alone: |expression| = k.
  2. Check k. If k < 0 → no solution (an absolute value is never negative).
  3. If k = 0 → exactly one case: expression = 0.
  4. If k > 0 → split into two cases: expression = k or expression = −k.
  5. Solve each equation separately.
  6. Check both solutions in the original — extra steps or a variable outside the |·| can introduce extraneous roots.

Special case — two absolute values: |A| = |B| → A = B or A = −B.

Solver — | a·x + b | = k

| ·x + | =

Example. |2x − 3| = 7 → 2x − 3 = 7 or 2x − 3 = −7 → x = 5 or x = −2.

The idea: distance from zerovisual

|x| = a asks "which points are exactly distance a from 0?" — for a > 0 there are two: x = a and x = −a.

Distance a from 0

BInequalities

1 · Multiplying / dividing by a numberflip demo

Multiply or divide both sides by a positive number → direction unchanged. By a negative number → direction flips.

Divide 2x < 6 by d

2 · Dividing by an expression of unknown signcase split

You cannot multiply/divide an inequality by x unless you know its sign. Fix — case split:

  1. Case 1: assume x > 0. Divide normally (keep direction). Combine with x > 0 using AND.
  2. Case 2: assume x < 0. Divide and flip. Combine with x < 0 using AND.
  3. Final answer = Case 1 OR Case 2 (x = 0 checked separately).

Alternative for rational inequalities: move everything to one side over a common denominator and build a sign chart — avoids dividing by the variable entirely.

Solver — 1 / x ▸ c

Worked example. 1/x > 2: Case 1 (x>0) → 1 > 2x → x < ½, AND x>0 → 0 < x < ½. Case 2 (x<0) → 1 < 2x → x > ½, AND x<0 → empty. Final: 0 < x < ½.

3 · Square roots in inequalitiesdomain + solver

Always start with the domain: whatever is under the root must be ≥ 0, combined with everything else using AND.

  • √(f(x)) < k: if k ≤ 0 → no solution. If k > 0 → f(x) ≥ 0 AND f(x) < k².
  • √(f(x)) > k: if k < 0 → all of the domain f(x) ≥ 0. If k ≥ 0 → f(x) > k² (domain automatic).
  • √A < √B → A ≥ 0 AND B ≥ 0 AND A < B.

Solver — √(x + p) ▸ k

√( x + )

Examples. √(x−2) < 3 → domain x ≥ 2, square x < 11, AND → 2 ≤ x < 11.   √(x+1) > 2 → x + 1 > 4 → x > 3.

4 · Absolute-value inequalitiesvisualizer
  • |f(x)| < k (k>0) → −k < f(x) < k — one bounded range, AND. If k ≤ 0 → no solution.
  • |f(x)| > k (k≥0) → f(x) < −k or f(x) > k — two disjoint rays. If k < 0 → true for all x.
  • ≤ / ≥ — same, with closed endpoints.

Visualizer — | x − c | ▸ k

Examples. |3x−1| < 5 → −4/3 < x < 2.   |3x−1| ≥ 5 → x ≤ −4/3 or x ≥ 2.

CWhen to use "and" vs "or"

Quick referencerule

Use AND when every condition must hold at the same point and the result is one intersected region: a domain restriction with a derived inequality, the double inequality from |f(x)| < k, or a case assumption (x > 0) with the inequality solved under it.

Use OR when the solution is separate alternatives, any one enough on its own: the two cases of an absolute-value equation, the two branches of |f(x)| > k, or merging the results of separate case-split branches.

Interactive — intersect (AND) vs union (OR)two intervals

Set two intervals and see how AND (overlap only) and OR (everything covered) differ.

Two intervals on the line

AND (intersection)– OR (union)–

1Angles from parallel lines & a transversal

The angle pairsvisualizer
  • Corresponding — same position at each crossing → equal.
  • Alternate interior — opposite sides of the transversal, between the lines → equal.
  • Alternate exterior — opposite sides, outside the lines → equal.
  • Co-interior (same-side interior) → supplementary, a + b = 180°.
  • Vertical angles (at one crossing) → always equal.

Drag the transversal angle

acute angles– obtuse angles– co-interior sum–

Converse: if any of these equalities holds, the two lines are parallel — the standard way to prove parallelism.

2Triangles

Triangle solver — 3 sidesinteractive

Angle sum = 180°. Triangle inequality: each side < sum of the other two. Largest side faces the largest angle.

Set the three sides

Every-triangle rulesreference
  • Exterior angle = sum of the two remote interior angles; exterior angles sum to 360°.
  • Midsegment (joins two midpoints) — parallel to the third side, half its length.
  • Centroid (medians) divides each median 2:1. Orthocenter = altitudes. Circumcenter = perpendicular bisectors. Incenter = angle bisectors.
  • Area: ½·base·height = ½·a·b·sin C = √(s(s−a)(s−b)(s−c)) (Heron, s = (a+b+c)/2).
Congruence & similarityreference

Congruent if: SSS, SAS, ASA, AAS, or HL (right triangles). SSA is not valid — the ambiguous case.

Similar if: AA, SSS~, or SAS~. Corresponding sides proportional; ratio of areas = (ratio of sides)².

Triangle types — sketch & rulesgallery

Pick a type — the sketch on the left, its attributes & formulas on the right.

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    3Quadrilaterals & polygons

    Polygon angle calculatorinteractive

    Convex n-gon: interior sum (n−2)·180°; exterior sum always 360°; diagonals n(n−3)/2.

    Number of sides

    interior angle sum– each interior (regular)– each exterior (regular)– diagonals–
    Quadrilateral types — sketch & rulesgallery

    Interior angles of any quadrilateral sum to 360°. Pick a type:

    –

      4Circles

      Inscribed vs central angleinteractive

      A central angle equals its arc. An inscribed angle is half the central angle on the same arc — so all inscribed angles on one arc are equal, and an angle in a semicircle is 90°.

      Change the arc

      central angle ∠AOB– inscribed angle ∠APB–
      Chordsreference
      • A radius/diameter perpendicular to a chord bisects it (and its arc).
      • Equal chords are equidistant from the center; longer chords sit closer to the center.
      • Intersecting chords: AP·PB = CP·PD.
      Tangents & power of a pointreference
      • A tangent is perpendicular to the radius at the point of tangency.
      • Two tangents from one external point are equal; the line to the center bisects the angle between them.
      • Tangent–chord angle = half the intercepted arc.
      • Two secants: PA·PB = PC·PD.   Tangent–secant: PT² = PA·PB.
      Cyclic quadrilateralreference

      A quadrilateral inscribed in a circle: opposite angles are supplementary (∠A + ∠C = 180°). An exterior angle equals the opposite interior angle.

      5Thales' theorem

      Thales' theorem — every formgallery

      Pick a form — the sketch on the left, its rules & formulas on the right. Drag the slider to see the ratio hold.

      –

        –
        Quick recap — the four formsreference
        • Small Thales (segment ratio): DE ∥ BC ⇒ AD/DB = AE/EC
        • Big Thales (whole-side ratio, △ADE ~ △ABC): AD/AB = AE/AC = DE/BC
        • Sand-clock Thales (hourglass through a point P): AB ∥ DE ⇒ PA/PD = PB/PE = AB/DE
        • Intercept theorem: parallel lines cut two transversals proportionally: AB/BC = A'B'/B'C'
        • Converse (all forms): equal ratios ⇒ the lines are parallel

        6Trigonometric geometry

        Right triangle — SOH-CAH-TOAinteractive
        • SOH: sin θ = opposite / hypotenuse
        • CAH: cos θ = adjacent / hypotenuse
        • TOA: tan θ = opposite / adjacent = sin θ / cos θ

        Drag the angle

        opposite (hyp = 1)– adjacent– sin θ– cos θ– tan θ–
        Right-triangle factsreference
        • Pythagoras: a² + b² = c² (c = hypotenuse)
        • Altitude to the hypotenuse: h² = p·q; each leg is the geometric mean of the hypotenuse and its adjacent segment
        • Median to the hypotenuse = c/2 = R (circumradius)
        • Special: 45-45-90 → 1 : 1 : √2,   30-60-90 → 1 : √3 : 2
        Law of Sines & Law of Cosinesinteractive + reference
        • Law of Sines: a/sin A = b/sin B = c/sin C = 2R  (R = circumradius)
        • Law of Cosines: c² = a² + b² − 2ab·cos C
        • rearranged for an angle: cos C = (a² + b² − c²)/(2ab)
        • Area: ½·a·b·sin C  or  (a²·sin B·sin C)/(2 sin A)

        Every triangle has one circumcircle through all three vertices, centre O, radius R. The Law of Sines just says each side over the sine of its opposite angle equals that circle's diameter:

        Drag angle A (moves B and C along the circle)

        R (circumradius)– side a = BC– angle A– a / sin A– 2R–

        Angle A is the inscribed angle on arc BC, so it's half that arc — while a/sin A stays pinned to 2R.

        SSA — the ambiguous casereference

        Two sides and a non-included angle can give 0, 1, or 2 triangles. After solving with the Law of Sines, check each candidate: angles must sum to 180° and all sides stay positive.

        Which tool for a triangle?decision helper

        Pick what you're given:

        1Powers

        Rules of powersreference
        • aᵐ·aⁿ = aᵐ⁺ⁿ,  aᵐ/aⁿ = aᵐ⁻ⁿ,  (aᵐ)ⁿ = aᵐⁿ
        • (ab)ⁿ = aⁿbⁿ,  (a/b)ⁿ = aⁿ/bⁿ
        • a⁰ = 1,  a⁻ⁿ = 1/aⁿ  (a ≠ 0)
        • a^(1/n) = ⁿ√a,  a^(m/n) = ⁿ√(aᵐ). If n is even, need a ≥ 0.
        Solving aˣ = aʸ — match the basesolver

        Rewrite both sides as powers of one base, then equate the exponents (valid since aˣ is one-to-one for a > 0, a ≠ 1).

        Solver — L^(p·x+q) = R^(r·x+s)

        ^( x + ) = ^( x + )

        Example. 4^(x+1) = 8^(x−2) → 2^(2x+2) = 2^(3x−6) → 2x+2 = 3x−6 → x = 8.

        Solving by substitution t = aˣmethod
        1. Set t = aˣ; rewrite a²ˣ → t², a⁻ˣ → 1/t, etc.
        2. Solve for t (often a quadratic).
        3. Reject any t ≤ 0 — aˣ > 0 always.
        4. Back-substitute aˣ = t and solve for x.

        Example. 4ˣ − 6·2ˣ + 8 = 0, t = 2ˣ → t² − 6t + 8 = 0 → t = 2, 4 → x = 1, 2.

        Power inequalities — flip by baseinteractive

        If a > 1: aˣ < aʸ ⟺ x < y (keep). If 0 < a < 1: aˣ < aʸ ⟺ x > y (flip).

        The function y = aˣ

        Examples. (1/2)ˣ < (1/2)³ → x > 3.   3ˣ > 3⁵ → x > 5.

        2Logs

        Definition, rules, identitiesreference
        • Definition: log_a(b) = c ⟺ aᶜ = b  (a > 0, a ≠ 1, b > 0)
        • log_a(1) = 0,  log_a(a) = 1
        • log_a(MN) = log_a M + log_a N,  log_a(M/N) = log_a M − log_a N
        • log_a(Mᵏ) = k·log_a M
        • Change of base: log_a(b) = log_c(b)/log_c(a),  log_a(b) = 1/log_b(a)
        • Inverse: a^(log_a b) = b,  log_a(aᵏ) = k

        Turn a number into a log: k = log_a(aᵏ) — e.g. 2 = log_3(9), so it can merge with other log_3 terms.

        Absolute value inside logs — log_a(x²)domain

        log_a(x²) is defined for every x ≠ 0. Writing it as 2·log_a(x) silently loses x < 0. The fix: log_a(x²) = 2·log_a|x|.

        Rule of thumb: pulling an even exponent out of a log ⇒ wrap the base in |·|. Odd exponent ⇒ not needed.

        Solving log_a(x + p) = csolver
        1. Domain: argument > 0.
        2. Definition: argument = aᶜ.
        3. Solve, then check against the domain.

        Solver — log_b(x + p) = c

        log ( x +) =

        Example. log_2(x − 1) = 3 → x − 1 = 8 → x = 9 (9 > 1 ✓).

        Solving log_a(b) = log_a(c)method
        1. Domain: b > 0 AND c > 0.
        2. log_a is one-to-one → equate arguments: b = c.
        3. Solve, then check the domain.

        Example. log(x+3) = log(2x−1) → domain x > 1/2 → x + 3 = 2x − 1 → x = 4 ✓.

        Log inequalitiessolver

        Two conditions with AND: the domain (every argument > 0) and the direction rule by base.

        • a > 1: log_a b < log_a c ⟺ 0 < b < c.
        • 0 < a < 1: log_a b < log_a c ⟺ b > c > 0.

        Solver — log_b(x + p) ▸ log_b(k)

        log ( x +) log2( )

        Example. log_2(x−1) < log_2(5) → x > 1 AND x < 6 → 1 < x < 6.

        0Domain of definition

        Risky forms & their conditionsreference
        FormDomain requirement
        1/f(x)f(x) ≠ 0
        x⁻ⁿx ≠ 0
        √f(x), ⁿ√f(x) (n even)f(x) ≥ 0
        aˣalways > 0 (never 0, never negative)
        log_b(a)a > 0 and b > 0, b ≠ 1
        tan(f(x))cos(f(x)) ≠ 0

        Write each condition, then intersect (AND) them all. A root in a denominator tightens ≥ 0 to > 0.

        Domain builderinteractive

        Add the conditions your expression forces; the intersection is the domain.

        Intersect conditions

        –

        Example. ln(x−1) / √(5−x) → x > 1 AND x < 5 → domain 1 < x < 5.

        1Derivatives

        Secant → tangentinteractive

        f'(x) = lim(h→0) [f(x+h) − f(x)] / h — the secant slope collapses to the tangent slope.

        Shrink h toward 0  (f(x) = ¼x², at x = 2)

        secant slope– true f'(2)1.000
        Differentiation rulesreference
        • d/dx[xⁿ] = n·xⁿ⁻¹,  d/dx[c] = 0,  d/dx[c·f] = c·f'
        • d/dx[f ± g] = f' ± g'
        • Product: d/dx[f·g] = f'g + fg'
        • Quotient: d/dx[f/g] = (f'g − fg') / g²
        • Chain: d/dx[f(g(x))] = f'(g(x))·g'(x)
        • Roots: convert first — √x = x^(1/2) → 1/(2√x).

        Examples. d/dx[(3x²+1)⁵] = 30x(3x²+1)⁴.   d/dx[√(x²+1)] = x/√(x²+1).

        Derivative playgroundinteractive

        f and f′, with the tangent line

        f(x)– f′(x) = tangent slope–
        Increasing / decreasing & sign chartsreference

        f increases where f'(x) > 0, decreases where f'(x) < 0. Critical points: f'(x) = 0 or undefined. Build a sign chart of f' with those as boundaries and test each interval.

        2Integrals

        Definite integral & signed areainteractive

        If F' = f then ∫f dx = F + C, and ∫[a,b] f dx = F(b) − F(a) (FTC) = signed area (below the axis counts negative).

        Area under a curve

        ∫[a,b] f dx (signed)– geometric area (|·| pieces)–
        Geometric area when part dips belowmethod
        1. Find where f(x) = 0 inside [a,b] — these split it into constant-sign pieces.
        2. Integrate each piece separately.
        3. Take the absolute value of any negative piece.
        4. Sum: total area = Σ |∫ over each sub-interval|.
        u-substitution / reverse chain rulemethod
        1. Let t = the inner expression; dt = (inner)' dx → isolate dx.
        2. Substitute so everything is in t.
        3. Integrate w.r.t. t, then substitute back.

        Shortcut: ∫ g'(x)·[g(x)]ⁿ dx = [g(x)]ⁿ⁺¹/(n+1) + C — undoing the chain rule.

        Example. ∫2x(x²+1)⁴ dx, t = x²+1 → ∫t⁴ dt = (x²+1)⁵/5 + C.

        3Trigonometric derivatives & integrals

        Derivativesreference
        • d/dx[sin x] = cos x
        • d/dx[cos x] = −sin x
        • d/dx[tan x] = 1/cos²x

        Chain versions (argument g(x)):

        • d/dx[sin(g)] = cos(g)·g'
        • d/dx[cos(g)] = −sin(g)·g'
        • d/dx[tan(g)] = g'/cos²(g)

        Example. d/dx[sin(3x²)] = 6x·cos(3x²).

        Integralsreference
        • ∫sin x dx = −cos x + C
        • ∫cos x dx = sin x + C
        • ∫1/cos²x dx = tan x + C

        For an inner g(x): ∫g'(x)·cos(g(x)) dx = sin(g(x)) + C, etc.

        4Exponential & logarithmic calculus

        Derivatives & integralsreference
        • d/dx[eˣ] = eˣ,  d/dx[aˣ] = aˣ·ln a
        • d/dx[ln x] = 1/x,  d/dx[log_a x] = 1/(x·ln a)
        • Chain: d/dx[e^g] = e^g·g',  d/dx[ln g] = g'/g
        • ∫eˣ dx = eˣ + C,  ∫aˣ dx = aˣ/ln a + C
        • ∫(1/x) dx = ln|x| + C — the |·| matches the full domain x ≠ 0.
        The number einteractive

        e = lim(n→∞) (1 + 1/n)ⁿ ≈ 2.71828 — the unique base where the tangent to y = aˣ at x = 0 has slope exactly 1, which is why eˣ is its own derivative.

        Watch (1 + 1/n)ⁿ approach e

        (1 + 1/n)ⁿ– e2.718281828 difference–

        Generated from the ETI/Math/base notes — Trig, Basic Algebra, Geometry, Powers & Logs, and Precalc. Fully offline & self-contained.