Look for this. A rotation described by distance along a circle has a natural unit, the radius. One radian sweeps an arc as long as the radius.
Use θ = s/r with matching lengths and assign a negative angle to clockwise motion. In standard position begin on the positive x-axis. Add or subtract full turns of 2π to locate the terminal arm. To convert to degrees, multiply by 180°/π.
Worked example
A pointer 12 cm long starts on the positive x-axis and sweeps 18π cm clockwise. Find its signed angle in radians and degrees and locate its tip.
The rotation is −18π/12 = −3π/2 or −270°. Adding 360° gives 90°, so the tip is 12 cm up the positive y-axis.
Try it yourself
A wheel of radius 0.4 m rolls straight ahead 3π m without slipping. It turns clockwise. Find the number of revolutions and the standard-position angle in [0, 2π) coterminal with that rotation.
Check your reasoning
The circumference is 0.8π m. The distance represents 3π/(0.8π) = 3.75 revolutions. The signed angle is −7.5π. Add four full turns (8π) to get π/2, whose terminal arm points up.
Outcome T2
Recover a missing unit-circle coordinate
Look for this. A unit-circle point is one unit from the origin. The stated quadrant picks the sign of a missing coordinate.
Substitute the known coordinate into x² + y² = 1. Take both square roots, then use the quadrant to select one. In standard position x = cos θ and y = sin θ.
Worked example
The terminal arm of θ meets the unit circle in quadrant III at (−5/13, y). Find y and tan θ exactly.
y² = 1 − (5/13)² = 144/169. Both coordinates are negative in quadrant III, so y = −12/13. Dividing y by x gives tan θ = 12/5.
Try it yourself
A proposed quadrant-II unit-circle point is (−0.6, 0.7). Test it, then keep its x-coordinate and correct its y-coordinate. What is cos θ there?
Check your reasoning
The sum of squares is 0.36 + 0.49 = 0.85, less than 1. For x = −0.6, y² = 0.64, so quadrant II gives y = +0.8. The corrected point has cos θ = −0.6.
Outcome T3
Build six ratios from one triangle
Look for this. A point on a terminal arm gives signed horizontal and vertical distances. They determine all six ratios even when the point lies outside the unit circle.
Find r = √(x² + y²). Use sin θ = y/r, cos θ = x/r and tan θ = y/x. Invert these ratios for csc, sec and cot respectively, checking for a zero divisor.
Worked example
An arm in standard position goes through (−20, 21). Determine all six ratios without estimating the angle.
The radius is √(400 + 441) = 29. Thus sin θ = 21/29, cos θ = −20/29 and tan θ = −21/20. Their reciprocals are csc θ = 29/21, sec θ = −29/20 and cot θ = −20/21.
Try it yourself
A 10 m cable runs from an anchor at the origin to a stage light at angle 5π/6 in standard position. Find its signed horizontal and vertical displacements and give sec(5π/6).
Check your reasoning
The angle is 150°, with a 30° reference angle. Multiplying cos θ = −√3/2 and sin θ = ½ by 10 gives −5√3 m horizontally and 5 m vertically. The reciprocal of cosine is sec θ = −2/√3.
Outcome T5
Use graph intersections to check an algebraic solution
Look for this. A quadratic in a single trig ratio can be factored. Each valid ratio can give several angles on the interval.
Rearrange to zero and factor in the trig ratio. Find all angles for each factor, respecting endpoints. On a graph of the original two sides, each intersection’s horizontal coordinate must be one of your solutions.
Worked example
Start with a graph to solve 2 sin²θ = 0.7 sin θ on 0 ≤ θ < 2π. Report each solution to two decimal places and use algebra to check that the list is complete.
Use radian mode. Set x from 0 to 2π and y from −1 to 2.2. Graph y = 2 sin²x and y = 0.7 sin x. Inspect the endpoint x = 0 and use the intersection tool near x = 0.36, 2.78 and 3.14. Exclude the right endpoint. Factoring sin θ(2 sin θ − 0.7) = 0 gives sin θ = 0 or 0.35. The first gives 0 and π. The second gives arcsin(0.35) and π − arcsin(0.35). The four solutions in order are approximately 0.00, 0.36, 2.78 and 3.14 radians.
Try it yourself
On −180° ≤ θ ≤ 180°, solve cos(2θ) = cos θ. Explain why a graph can show a solution even when the curves just touch.
Check your reasoning
Substitute cos(2θ) = 2 cos²θ − 1. Then (2 cos θ + 1)(cos θ − 1) = 0. Cosine equals −½ at −120° and 120°, and 1 at 0°. In degree mode both curves reach height 1 at 0° and turn back there. Touching still gives equal outputs.
Worked notes and diagrams
1
Angles, named
The idea: trigonometry has more vocabulary than mathematics, and questions assume all of it.
Word
What it means
Angle in standard position
Drawn with its corner at the origin and one arm along the positive x-axis.
Initial arm
The arm you start from, always the positive x-axis.
Terminal arm
The arm you finish at, after turning through the angle.
Positive angle
A turn counterclockwise. A negative angle turns clockwise.
Quadrant
One of the four regions the axes cut the plane into, numbered I to IV counterclockwise from the upper right.
Reference angle
The acute angle between the terminal arm and the nearest part of the x-axis — always positive.
Coterminal angles
Different angles whose terminal arms land in the same place: 30°, 390° and −330°.
Radian
The angle that wraps one radius of arc around the circle. A full turn is 2π radians.
Arc
A piece of the circle's edge. Its length is a = θr, with θ in radians.
Unit circle
A circle of radius 1 centred at the origin, where the terminal point is (cos θ, sin θ).
General solution
All the angles that satisfy an equation. Include every branch and its repeats, using integer n. For example, tan θ = 1 gives θ = π/4 + nπ.
2
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.
3
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.
CAST tells you which ratios are positive: All in Quadrant I, Sine in II, Tangent in III, Cosine in IV.
Try it: drag around the unit circle
Drag the pink point, tap anywhere on the circle, or use the slider. Watch how the signs change as you cross each axis, and how the special angles give exact values.
Exact values worth knowing
θ
sin θ
cos θ
tan θ
30° = π/6
1/2
√3/2
1/√3
45° = π/4
√2/2
√2/2
1
60° = π/3
√3/2
1/2
√3
90° = π/2
1
0
undefined
Any angle in three steps
Find cos(4π/3):
Quadrant: 4π/3 is past π, so it is in Quadrant III.
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.
4
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.
For y = sin x, a = 1, b = 1, c = 0 and d = 0: amplitude 1, period 2π, midline y = 0.
Parameter
What it changes
How to read it
a
Amplitude
|a|, the distance from midline to peak
b
Period
2π ÷ |b| (or 360° ÷ |b|)
c
Horizontal shift
Right by c if written as (x − c)
d
Midline
y = d; max is d + |a|, min is d − |a|
Try it: build a sinusoid
Move one slider at a time and watch which part of the graph changes. Then try a match challenge: recreate the dashed target curve.
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.
5
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.
Reference angle: sin(π/6) = 1/2, so it is π/6.
Sine is positive in Quadrants I and II.
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.
Remove a double angle with an identity
For cos 2θ = sin θ on [0, 2π), substitute cos 2θ = 1 − 2sin²θ. Rearrange to (2sin θ − 1)(sin θ + 1) = 0. Sine equals ½ at π/6 and 5π/6, and −1 at 3π/2. These three angles satisfy the original equation.
Identities to know
Reciprocal: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ where both sides are defined. For cot θ throughout its domain, use cos θ/sin θ with sin θ ≠ 0.
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. Keep the original restrictions sin x ≠ 0 and cos x ≠ 0.
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.
Practice by topic
Work on paper, choose an answer, then compare your method with the explanation.
Angles and radians · 4 questions
Practice 1
What is 180° in radians?
π
2π
π/2
180π
Answer and explanation
π
Half a turn is π radians, so 180° = π. That one fact gives every conversion: multiply degrees by π/180.
Practice 2
What is 3π/4 in degrees?
120°
135°
150°
270°
Answer and explanation
135°
Replace π with 180°: 3 × 180° ÷ 4 = 135°.
Practice 3
Which angle is coterminal with 45°?
135°
−45°
405°
225°
Answer and explanation
405°
Coterminal angles differ by full turns: 45° + 360° = 405°. −45° is a reflection across the x-axis, not the same arm.
Practice 4
A circle has radius 6 cm. What arc length does a central angle of 2π/3 cut off?
4π cm
720 cm
2π cm
12π cm
Answer and explanation
4π cm
Arc length is a = rθ with θ in radians: 6 × 2π/3 = 4π cm. Using 120° instead of radians gives the nonsense 720.
The unit circle · 4 questions
Practice 1
What is the exact value of sin 30°?
√2/2
1
1/2
√3/2
Answer and explanation
1/2
In the 30-60-90 triangle the side opposite 30° is half the hypotenuse, so sin 30° = 1/2. √3/2 is sin 60°.
Practice 2
An angle in standard position has sin θ > 0 and cos θ < 0. Which quadrant is its terminal arm in?
Quadrant III
Quadrant II
Quadrant I
Quadrant IV
Answer and explanation
Quadrant II
Sine follows y and cosine follows x. Positive y with negative x is Quadrant II — the S in CAST, where only sine is positive.
Practice 3
The point P(−3, 4) lies on the terminal arm of angle θ. What is cos θ?
3/5
4/5
−3/5
−4/3
Answer and explanation
−3/5
First find r = √((−3)² + 4²) = 5. Then cos θ = x/r = −3/5. The sign comes from x, which is negative in Quadrant II.
Practice 4
What is the exact value of cos(4π/3)?
1/2
−√3/2
−1/2
√3/2
Answer and explanation
−1/2
4π/3 is in Quadrant III with reference angle π/3. cos(π/3) = 1/2, and cosine is negative in Quadrant III, so −1/2.
Sinusoidal graphs · 5 questions
Practice 1
What is the amplitude of y = 3 sin x?
2π
1/3
6
3
Answer and explanation
3
The amplitude is |a|, the distance from the midline to a peak: 3. The full height from trough to peak would be 6.
Practice 2
What is the period of y = sin x, in radians?
π
π/2
4π
2π
Answer and explanation
2π
One full turn around the unit circle brings sine back to where it started, and a full turn is 2π.
Practice 3
What is the period of y = 2 cos(3x), in radians?
2π/3
6π
3π
2π
Answer and explanation
2π/3
The period is 2π ÷ |b| = 2π/3. The 2 in front changes the amplitude, never the period.
Practice 4
What is the maximum value of y = 2 sin x + 5?
5
10
7
2
Answer and explanation
7
The midline is y = 5 and the amplitude is 2, so the graph peaks at 5 + 2 = 7 and bottoms out at 3.
Practice 5
What is the range of y = 3 sin[2(x − π/4)] − 1?
−4 ≤ y ≤ 2
−3 ≤ y ≤ 3
−2 ≤ y ≤ 4
−1 ≤ y ≤ 5
Answer and explanation
−4 ≤ y ≤ 2
Only the amplitude (3) and vertical shift (−1) affect the range: from −1 − 3 = −4 up to −1 + 3 = 2. The 2 and π/4 move the graph sideways.
Equations and identities · 3 questions
Practice 1
Solve sin θ = 1/2 for 0 ≤ θ < 2π.
π/6 and 5π/6
π/3 and 2π/3
π/6 only
π/6 and 7π/6
Answer and explanation
π/6 and 5π/6
The reference angle is π/6, and sine is positive in Quadrants I and II: π/6 and π − π/6 = 5π/6. Stopping at one answer is the usual slip.
Practice 2
Solve 2cos²θ − cos θ − 1 = 0 for 0 ≤ θ < 2π.
0, 2π/3, 4π/3
2π/3, 4π/3
0, π/3, 5π/3
π, π/3, 5π/3
Answer and explanation
0, 2π/3, 4π/3
Treat it as a quadratic in cos θ: (2cos θ + 1)(cos θ − 1) = 0. cos θ = −1/2 gives 2π/3 and 4π/3; cos θ = 1 gives 0. Dropping the 0 is the common miss.
Practice 3
How many solutions does cos 2θ = cos θ have for 0 ≤ θ < 2π?
3
6
2
4
Answer and explanation
3
Replace cos 2θ by 2cos²θ − 1, then factor (2cos θ + 1)(cos θ − 1) = 0. The solutions are 2π/3, 4π/3 and 0. The endpoint 2π is excluded.
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