SME
Grade 11 Math · Algebra 2
The unit circle and radian measure
Use this for “Find the point on the unit circle at 5π/6, and its sine, cosine and tangent.”
Diagram
θ
=
5π/6:
the
point
(cos
θ,
sin
θ)
is
(−
√
3
2
,
1
2
).
Reference angle: π/6.
tan
θ
=
sin θ
cos θ
=
−
√
3
3
.
This is an example. Type your own values to start a new problem.
(cos
θ
,
sin
θ
)
=
,
1
2
θ°
Angle in degrees
worked out (not typed)
m
Tangent, tan θ (the slope y ÷ x)
worked out (not typed) · ≈ −0.5774
Formulas
θ
°
=
180
×
θ
π
150 = 180 × 5π/6 ÷ π
x
=
cos(
θ
)
−
√
3
2
=
cos(5π/6)
y
=
sin(
θ
)
1
2
=
sin(5π/6)
m
=
y
x
−
√
3
3
=
1
2
÷
(−
√
3
2
)
Assumptions
•
The point at angle θ is (cos θ, sin θ), because the radius is 1.
•
The reference angle and the quadrant give the value and its sign.
•
One turn is 2π radians, 360°.
Step-by-step
WE KNOW
θ = 5π/6
FIND
angle in degrees (θ°), cos θ (x), sin θ (y), tangent, tan θ (the slope y ÷ x) (m)
Step 1 · Find angle in degrees (θ°)
Use θ° = 180 × θ ÷ π
π radians is 180°.
θ
°
=
180
×
5π/6
÷
π
θ
°
=
180
×
5
6
θ
°
=
150
Step 2 · Find cos θ (x)
Use x = cos(θ)
The
x
-coordinate
of
the
point
on
the
circle.
x
=
cos(5π/6)
x
=
−
√
3
2
≈
−0.866
Step 3 · Find sin θ (y)
Use y = sin(θ)
The
y
-coordinate
of
the
point
on
the
circle.
y
=
sin(5π/6)
y
=
1
2
Step 4 · Find tangent, tan θ (the slope y ÷ x): m
Use m = y ÷ x
The tangent is the sine over the cosine: rise over run.
m
=
1
2
÷
(−
√
3
2
)
m
=
1
2
÷
((−1.732)/2)
m
=
1
2
÷
(−0.866)
m
=
−
√
3
3
≈
−0.5774
Check
Put the numbers back in. Both sides should match.
150 = 180 × 5π/6 ÷ π ✓
−
√
3
2
=
cos(5π/6)
✓
1
2
=
sin(5π/6)
✓
−
√
3
3
=
1
2
÷
(−
√
3
2
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Arc length, sector area and radian measure
The Pythagorean theorem's converse and special right triangles
More problem types
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A point on the terminal side
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