SME
Grade 11 Math · Algebra 2
Pythagorean trigonometric identities
Use this for “sin θ = 0.6 and θ is in quadrant II. Find cos θ and tan θ.”
Diagram
θ = 143.13°: the point (cos θ, sin θ) is ≈ (−0.8, 0.6).
Reference angle: 36.87°.
tan
θ
=
sin θ
cos θ
≈
−0.75.
Quadrant: 2
This is an example. Type your own values to start a new problem.
(
)
2
+
−4
5
2
=
1
Q
Quadrant
example
m
Tangent, tan θ
worked out (not typed)
θ
Angle
worked out (not typed) · °
Formulas
c
=
±
√
1 −
s
2
,
the
sign
from
quadrant
Q
−
4
5
=
±√(1
−
(
3
5
)
2
),
the
sign
from
quadrant
2
m
=
s
c
−
3
4
=
3
5
÷
(−
4
5
)
θ
=
sin⁻¹(
s
),
moved
into
quadrant
Q
143.1301
=
sin⁻¹(
3
5
),
moved
into
quadrant
2
Assumptions
•
The point (cos θ, sin θ) is on the circle x² + y² = 1, so sin²θ + cos²θ = 1.
•
The square root gives the size; the quadrant gives the sign.
•
It holds for every angle, 3x as well as θ.
Step-by-step
WE KNOW
s = 3/5, Q = 2
FIND
cosine, cos θ (c), tangent, tan θ (m), angle (θ)
Step 1 · Find cosine, cos θ (c)
Use c = ±√(1 − s²), the sign from quadrant Q
cos²
θ
=
1
−
sin²
θ
;
cosine
is
negative
in
quadrants
II
and
III.
c
=
−
√
1 −
s
2
c
=
−√(1
−
(
3
5
)
2
)
c
=
−
√
1 −
9
25
c
=
−
√
16
25
c
=
−
4
5
Step 2 · Find tangent, tan θ (m)
Use m = s ÷ c
Tangent is sine over cosine.
m
=
3
5
÷
(−
4
5
)
m
=
−
3
4
Step 3 · Find angle (θ)
Use θ = sin⁻¹(s), moved into quadrant Q
Mirror the inverse sine across the y-axis into quadrants II and III.
θ
=
180
−
sin⁻¹(
s
)
θ
=
180
−
sin⁻¹(
3
5
)
θ
=
180°
−
36.8699°
θ
=
143.1301°
Check
Put the numbers back in. Both sides should match.
−
4
5
=
−√(1
−
(
3
5
)
2
)
✓
−
3
4
=
3
5
÷
(−
4
5
)
✓
143.1301
=
180
−
sin⁻¹(
3
5
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.