SME
Grade 12 Math · Problem type
Half-angle formulas
Use this for “cos A = 7/25 and A/2 is in Quadrant I. Find sin(A/2) and cos(A/2).”
Diagram
θ = 36.87°: the point (cos θ, sin θ) is ≈ (0.8, 0.6).
This is an example. Type your own values to start a new problem.
sin(A/2)
=
±
1
−
2
=
3
5
S
cos(A/2)
=
±
1
+
2
=
4
5
K
Q
Quadrant of A/2
example
H
The half angle, A/2
worked out (not typed) · °
Formulas
S
=
±√(
1 −
c
2
),
the
sign
from
quadrant
Q
3
5
=
±√(
1 −
7
25
2
),
the
sign
from
quadrant
1
K
=
±√(
1 +
c
2
),
the
sign
from
quadrant
Q
4
5
=
±√(
1 +
7
25
2
),
the
sign
from
quadrant
1
H
=
the
angle
in
quadrant
Q
with
cosine
K
36.8699
=
the
angle
in
quadrant
1
with
cosine
4
5
Assumptions
•
sin(A/2) = ±√((1 − cos A) ÷ 2) and cos(A/2) = ±√((1 + cos A) ÷ 2).
•
The signs come from the quadrant of A/2, not of A: the sine is positive in I and II, the cosine in I and IV.
•
For A from 0° to 360°, A/2 is in quadrant I or II; a larger or negative A can put it in III or IV.
Step-by-step
WE KNOW
c = 7/25, Q = 1
FIND
sin(A/2) (S), cos(A/2) (K), the half angle, A/2 (H)
Step 1 · Find sin(A/2): S
Use S = ±√((1 − c) ÷ 2), the sign from quadrant Q
The
half-angle
formula
for
sine;
the
root
is
negative
when
A
2
is
in
quadrant
III
or
IV.
S
=
√(
1 −
c
2
)
S
=
√(
1 −
7
25
2
)
S
=
√(
0.72
2
)
S
=
√
0.36
S
=
3
5
Step 2 · Find cos(A/2): K
Use K = ±√((1 + c) ÷ 2), the sign from quadrant Q
The
half-angle
formula
for
cosine;
the
root
is
negative
when
A
2
is
in
quadrant
II
or
III.
K
=
√(
1 +
c
2
)
K
=
√(
1 +
7
25
2
)
K
=
√(
1.28
2
)
K
=
√
0.64
K
=
4
5
Step 3 · Find the half angle, A/2 (H)
Use H = the angle in quadrant Q with cosine K
cos⁻¹ gives an angle from 0° to 180°; in quadrants III and IV take it from 360°.
H
=
cos⁻¹(
K
)
H
=
cos⁻¹(
4
5
)
H
=
36.8699°
Check
Put the numbers back in. Both sides should match.
3
5
=
√(
1 −
7
25
2
)
✓
4
5
=
√(
1 +
7
25
2
)
✓
cos(36.8699°)
=
4
5
✓
Topic check
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