01 · A negative angle
Find sin(−150°), cos(−150°) and tan(−150°).
Hint
Add 360° to get 210°.
Worked solution
sin = −1/2, cos = −√3/2, tan = √3/3
Understand · explore · practise
Unit circle and exact trig values
Understand sine, cosine and tangent for any angle. Use quadrants, reference angles and exact values, with a unit-circle model and worked practice.
Before you startRight-triangle trigonometry, Pythagoras and surds
01 / The unit circle
One point gives all three ratios.
A unit circle has radius 1 and centre O = (0,0). Measure θ anticlockwise from the positive x-axis; a clockwise turn is negative. The point P reached on the circle has coordinates:
P = (cos θ, sin θ)
tan θ = sin θ / cos θ, when cos θ ≠ 0
Cosine is the horizontal coordinate; sine is the vertical coordinate. Tangent is the gradient of OP, so it is undefined when OP is vertical. The angle may exceed one complete turn. Throughout these lessons angles are in degrees.
Signed coordinates on a unit circleChoose and compare
θ = 90°, equivalent to 90° and on an axis. P ≈ (0, 1). Tangent is undefined because cosine is zero.
Watch the coordinates change sign around the circle
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Quadrants and axes
The coordinate signs explain the signs of the ratios.
- Quadrant I, 0° < θ < 90°: sin, cos and tan are positive.
- Quadrant II, 90° < θ < 180°: sin is positive; cos and tan are negative.
- Quadrant III, 180° < θ < 270°: tan is positive; sin and cos are negative.
- Quadrant IV, 270° < θ < 360°: cos is positive; sin and tan are negative.
The axes are boundaries, not part of those open quadrants. At 90° and 270°, cosine is zero and tangent is undefined. At 0° and 180°, sine and tangent are zero. Zero is neither positive nor negative.
03 / Negative and large angles
Remove whole turns without changing the point.
sin(θ + 360°n) = sin θ
cos(θ + 360°n) = cos θ
tan(θ + 180°n) = tan θ
n is any integer; tangent identities apply where both sides are defined.
For 765°, subtract 720° to reach 45°. For −120°, add 360° to reach 240°. Reflection in the x-axis changes the vertical coordinate but preserves the horizontal one:
sin(−θ) = −sin θ
cos(−θ) = cos θ
tan(−θ) = −tan θ
Thus sin 765° = √2/2, cos(−120°) = −1/2 and tan(−120°) = √3. Tangent repeats after 180° because both coordinates change sign and their quotient stays the same.
04 / Reference angles
Use the acute angle to the x-axis, then attach the sign.
For an angle strictly inside a quadrant, its reference angle α is the acute angle between OP and the nearest part of the x-axis. First reduce θ to 0° ≤ θ < 360°.
QI: α = θ
QII: α = 180° − θ
QIII: α = θ − 180°
QIV: α = 360° − θ
For θ = 225°, α = 45°. The magnitudes are those of 45°, but both coordinates are negative: sin 225° = cos 225° = −√2/2, while tan 225° = 1. For 330°, α = 30°: sine is negative and cosine positive.
Related-angle identities
sin(180° − θ) = sin θ
cos(180° − θ) = −cos θ
sin(180° + θ) = −sin θ
cos(180° + θ) = −cos θ
sin(90° − θ) = cos θ
cos(90° − θ) = sin θ
These follow by reflecting or rotating the unit-circle point. They apply to general θ, not only acute θ. For tangent, form the quotient and check that its denominator is non-zero.
05 / Derive exact values
Two familiar triangles supply the important angles.
An isosceles right triangle with legs 1 and 1 has hypotenuse √2, giving sin 45° = cos 45° = 1/√2 = √2/2 and tan 45° = 1.
Bisect an equilateral triangle of side 2. Each right triangle has hypotenuse 2, short leg 1 and long leg √3. The short leg is opposite 30° and the long leg opposite 60°.
sin 30° = 1/2 · cos 30° = √3/2 · tan 30° = √3/3
sin 60° = √3/2 · cos 60° = 1/2 · tan 60° = √3
At 0°, P = (1,0); at 90°, P = (0,1). Thus sin 0° = 0, cos 0° = 1, tan 0° = 0; sin 90° = 1, cos 90° = 0 and tan 90° is undefined.
06 / Keep answers exact
Combine the reference angle with the quadrant sign.
Evaluate without a calculatorWorked example
sin 150° = sin 30° = 1/2
A positive vertical coordinate in quadrant II.
cos 240° = −cos 60° = −1/2
A negative horizontal coordinate in quadrant III.
tan 315° = −tan 45° = −1
The coordinates have opposite signs.
2 sin 150° − √3 cos 210° = 1 + 3/2 = 5/2
Since cos 210° = −√3/2, subtracting that term adds 3/2.
Do not replace exact surds with rounded decimals during the calculation. Write 1/√3 as √3/3 when a rational denominator is needed.
07 / Extension: 15°
Bisect a chord to find a new exact value.
On a unit circle take A = (1,0) and B = (√3/2,1/2), so angle AOB is 30°. Triangle OAB is isosceles. Its median OM to the midpoint M of AB is perpendicular to AB and bisects the angle at O.
M = ((2 + √3)/4, 1/4)
OM² = ((2 + √3)/4)² + (1/4)²
= (2 + √3)/4
In right triangle OMA, OA = 1 and angle AOM = 15°, so OM = cos 15°. Pythagoras then gives sin² 15° = 1 − OM². Both ratios are positive.
cos 15° = √(2 + √3)/2 = (√6 + √2)/4
sin 15° = √(2 − √3)/2 = (√6 − √2)/4
Check the alternative surd forms by squaring them. Their positive signs identify the correct square roots. This derivation uses geometry and Pythagoras without assuming a new angle formula.
08 / Extension: a pyramid
Use a right triangle inside each triangular face.
A right square pyramid has base side 6 and four equal sloping edges of length 5. Each triangular face has sides 5,5,6. Its perpendicular height to the 6-unit base bisects that base.
Face height = √(5² − 3²) = 4
One face area = ½ × 6 × 4 = 12
Total surface area = 6² + 4 × 12 = 84
The height of a triangular face is not the vertical height of the pyramid. The surface area uses four face areas and the square base.
If every edge has length s
Each face is equilateral. Its height is s√3/2, so its area is s²√3/4. Including the square base gives total surface area (1 + √3)s². The result is exact because the special-triangle height is exact.
09 / Your turn
Start with a point, a sign or a reference angle.
Give exact values. A calculator can check your result afterwards.
02 · More than a turn
Evaluate cos 840° and tan 585°.
Hint
Cosine repeats every 360°; tangent every 180°.
Worked solution
cos 840° = cos 120° = −1/2
tan 585° = tan 45° = 1
03 · Axis boundary
Find sin 270°, cos 270° and tan 270°.
Hint
The point is (0,−1).
Worked solution
sin 270° = −1, cos 270° = 0
tan 270° is undefined, not zero.
04 · Exact arithmetic
Evaluate 4 sin 330° + 2√3 cos 150°.
Hint
Both requested ratios are negative.
Worked solution
4(−1/2) + 2√3(−√3/2) = −2 − 3 = −5
05 · Reference angle
Find the reference angle and signs of all three ratios for θ = 1020°.
Hint
Subtract 720°.
Worked solution
θ is equivalent to 300°; reference angle = 60°.
sin negative, cos positive, tan negative.
06 · Same cosine
Explain geometrically why cos(360° − θ) = cos θ.
Hint
Reflect the point in the x-axis.
Worked solution
The reflected point has the same x-coordinate and opposite y-coordinate. Cosine is the x-coordinate, so it is unchanged.
07 · Complement
Evaluate sin 75° exactly using the 15° result.
Hint
sin(90° − θ) = cos θ.
Worked solution
sin 75° = cos 15° = (√6 + √2)/4
08 · Product signs
For 180° < θ < 270°, is sin θ cos θ positive or negative? Is sin θ + cos θ positive or negative?
Hint
Both coordinates are negative.
Worked solution
The product is positive; the sum is negative. Neither coordinate is zero inside this open quadrant.
10 / Recap
Coordinates make the signs predictable.
- P = (cos θ, sin θ) on the unit circle.
- Tangent is the gradient sin θ / cos θ and needs cos θ ≠ 0.
- Use whole turns and a reference angle before calculating.
- The axes need their own zero/undefined checks.
- Derive the 30°, 45° and 60° values from triangles, then keep them exact.
Section 1 of 10 · The unit circle