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Math 30-1 · Trigonometry

Angles, circles and waves

Trig in Math 30-1 is one picture — a circle with radius 1 — looked at four different ways. Get comfortable with the circle and everything else comes from it.

Radians

The idea: a radian measures an angle by arc length. Half a turn is π radians, so π = 180°. Everything else is a conversion from that.

To convert degrees to radians, multiply by π/180. To go back, multiply by 180/π — or just replace π with 180°.

  • 90° = π/2,   60° = π/3,   45° = π/4,   30° = π/6
  • 3π/4 = 3 × 180° ÷ 4 = 135°

Standard position, coterminal and reference angles

An angle in standard position starts on the positive x-axis and turns counterclockwise (clockwise if negative). Coterminal angles end on the same arm — they differ by full turns, so 45°, 405° and −315° are all coterminal. The reference angle is the acute angle between the terminal arm and the x-axis; 150°, 210° and 330° all have a reference angle of 30°.

Arc length

With θ in radians, arc length is a = rθ. A radius of 6 cm and an angle of 2π/3 gives a = 6 × 2π/3 = 4π cm.

Watch out: a = rθ only works in radians. Plugging in 120° gives 720 cm for a 6 cm circle, which is obviously too big.

The unit circle

The idea: on a circle of radius 1, the point at angle θ is (cos θ, sin θ). Cosine is the x-coordinate, sine is the y-coordinate, and tan θ = y/x.

(cos θ, sin θ) cos θ sin θ 1 θ SATC
CAST tells you which ratios are positive: All in Quadrant I, Sine in II, Tangent in III, Cosine in IV.

Exact values worth knowing

θsin θcos θtan θ
30° = π/61/2√3/21/√3
45° = π/4√2/2√2/21
60° = π/3√3/21/2√3
90° = π/210undefined

Any angle in three steps

Find cos(4π/3):

  1. Quadrant: 4π/3 is past π, so it is in Quadrant III.
  2. Reference angle: 4π/3 − π = π/3, and cos(π/3) = 1/2.
  3. Sign: only tangent is positive in III, so cosine is negative: −1/2.

A point that is not on the unit circle

If P(−3, 4) is on the terminal arm, first find r = √(x² + y²) = 5. Then sin θ = y/r = 4/5, cos θ = x/r = −3/5, and tan θ = y/x = −4/3.

Watch out: getting the number right but the sign wrong. Always decide the quadrant before you write the answer.

Sinusoidal graphs

The idea: y = a sin b(x − c) + d. Each letter does exactly one job: a stretches up and down, b squeezes sideways, c slides sideways, d slides up and down.

amplitude = |a| period = 2π ÷ |b| midline y = d
For y = sin x, a = 1, b = 1, c = 0 and d = 0: amplitude 1, period 2π, midline y = 0.
ParameterWhat it changesHow to read it
aAmplitude|a|, the distance from midline to peak
bPeriod2π ÷ |b| (or 360° ÷ |b|)
cHorizontal shiftRight by c if written as (x − c)
dMidliney = d; max is d + |a|, min is d − |a|

For y = 3 sin[2(x − π/4)] − 1: amplitude 3, period π, shifted π/4 right, midline y = −1. So the range is −4 ≤ y ≤ 2.

Watch out: y = sin(2x − π/2) is not shifted π/2. Factor first: sin 2(x − π/4), so the shift is π/4.

Solving equations and using identities

The idea: isolate the trig function, find the reference angle, then use CAST to find every angle in the domain — there is almost always more than one.

Worked example

Solve sin θ = 1/2 for 0 ≤ θ < 2π.

  1. Reference angle: sin(π/6) = 1/2, so it is π/6.
  2. Sine is positive in Quadrants I and II.
  3. Quadrant I: π/6. Quadrant II: π − π/6 = 5π/6.

When it looks like a quadratic

2cos²θ − cos θ − 1 = 0 factors exactly like 2x² − x − 1: (2cos θ + 1)(cos θ − 1) = 0. So cos θ = −1/2, giving 2π/3 and 4π/3, or cos θ = 1, giving 0. Three solutions.

Double angles mean double the solutions

sin 2θ = 1/2 has a period of π, so it runs through two full cycles on [0, 2π) and hits 1/2 four times. Solve for 2θ over [0, 4π), then divide by 2.

Identities to know

  • Reciprocal: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ
  • Quotient: tan θ = sin θ / cos θ
  • Pythagorean: sin²θ + cos²θ = 1
  • Sum and difference: sin(A ± B) = sin A cos B ± cos A sin B, and cos(A ± B) = cos A cos B ∓ sin A sin B
  • Double angle: sin 2A = 2 sin A cos A, and cos 2A = cos²A − sin²A

To simplify (1 − cos²x) ÷ (sin x cos x): the Pythagorean identity turns the top into sin²x, and sin²x ÷ (sin x cos x) = sin x ÷ cos x = tan x.

Watch out: dividing both sides by a trig function, like cos θ, to "cancel" it. You lose the solutions where cos θ = 0. Factor instead.

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