Cosine is the x-coordinate; sine is the y.
In a right triangle cos = adjacent/hypotenuse. With the hypotenuse equal to 1 the division disappears and the adjacent side simply is the cosine.
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- The Unit Circle and the Circular Functions
The Unit Circle and the Circular Functions
On a circle of radius 1, the point at angle θ is simply (cos θ, sin θ).
Find the value
cos(2π3)
2π3 is 120°, which lies in quadrant II with a reference angle of 60°. There cos 60° = 12, and cosine is negative in quadrant II, so the answer is −12. A forgets the sign, and C uses the reference angle's SINE instead of its cosine.
Three special angles carry almost everything.
| θ | 0 | π6 | π4 | π3 | π2 |
|---|---|---|---|---|---|
| sin θ | 0 | 12 | √22 | √32 | 1 |
| cos θ | 1 | √32 | √22 | 12 | 0 |
Read the sine row left to right as √02, √12, √22, √32, √42 and it is just the square roots of 0 to 4 over 2. The cosine row is the same list backwards.
The quadrant fixes the signs.
quadrant I
all positive
x and y both positive
quadrant II
sine positive only
x negative, y positive
III then IV
tangent, then cosine
"All Students Take Chemistry"
The signs are not a rule to memorise separately — they are just the signs of x and y in that quadrant. If you can picture where the point sits, you can read the signs off.
Reference angle for the size, quadrant for the sign.
5π4 = 225°
quadrant III
reference angle 225 − 180 = 45°
to the x-axis
sin 45° = √22
the size
sine is negative in III: −√22
the sign
Two separate questions, answered separately: how big, and what sign. Trying to do both at once is what produces the right number with the wrong sign.
Tangent at any angle: divide, then fix the sign.
Tangent is not in the table, but it is built from the two rows that are.
Find tan(5π6).
5π6 = 150°, quadrant II
reference angle 180 − 150 = 30°
tan 30° = (12) ÷ (√32) = 1√3 = √33
sin over cos, the 2s cancel
tangent is negative in II
y positive, x negative
tan(5π6) = −√33
size and sign together
Tangent is y ÷ x, so it is positive only where x and y share a sign: quadrants I and III.
Reading sin instead of tan
Stopping at sin 30° = 12 gives −12, which is not a tangent value here at all. For the 45° family tangent is ±1; for 30° it is ±√33; for 60° it is ±√3.
Variations: a negative angle, an angle past a full turn, and a point on an axis.
The same reference-angle method, after one extra move to find where the point sits.
A negative angle: cos(−3π4)
−3π4 = −135°; add 360°: 225°
quadrant III
reference 45°, cos 45° = √22
the size
cos(−3π4) = −√22
cosine is negative in III
Clockwise angles land somewhere too; find the coterminal positive angle first.
Past a full turn: cos(11π3)
11π3 − 2π = 11π3 − 6π3 = 5π3
remove one full turn
5π3 = 300°, quadrant IV, reference 60°
cos 60° = 12
cos(11π3) = 12
cosine is positive in IV
On an axis: csc(3π2)
3π2 is the bottom point (0, −1)
no reference angle needed
sin(3π2) = −1
the y-coordinate
csc(3π2) = 1−1 = −1
the reciprocal
Quadrantal angles are read straight off the four axis points. The reciprocal of ±1 is itself.
The other four functions come from these two.
tan θ
= sin θcos θ
undefined where cos = 0
csc θ, sec θ
= 1sin θ, 1cos θ
the reciprocals
cot θ
= cos θsin θ
undefined where sin = 0
Because x² + y² = 1 on the unit circle, sin²θ + cos²θ = 1 for every angle. That is the Pythagorean identity, and it is the circle's equation wearing different clothing — the next lesson is built on it.
On the exam
Tests ask for an exact value such as sin, cos or tan of a multiple of π6 or π4, or ask in which quadrant a function is negative. The fastest route: convert to degrees, find the quadrant and reference angle, take the size from the 30-45-60 table and the sign from the quadrant. A 10-second check: the size must be one of 0, 12, √22, √32 or 1 for sine and cosine, and the sign must match where the point sits.
Now you try
Practice problem
Find the value
sin(4π3)
4π3 is 240°, in quadrant III, with a reference angle of 240 − 180 = 60°. sin 60° = √32, and sine is negative in quadrant III, so the answer is −√32. D uses the reference angle's COSINE, and B loses the sign.
Hint: convert to degrees first if it helps, find the quadrant, then subtract 180° for the reference angle. Do the size and the sign as two separate steps.