SME
Grade 12 Math · Problem type
Double-angle formulas
Use this for “sin A = 3/5 with A in Quadrant II. Find sin 2A and cos 2A.”
Diagram
θ = 143.13°: the point (cos θ, sin θ) is ≈ (−0.8, 0.6).
Reference angle: 36.87°.
This is an example. Type your own values to start a new problem.
sin
2A
=
2(
s
)
−4
5
=
−24
25
S
cos
2A
=
−4
5
2
−
(
)
2
=
7
25
K
Q
Quadrant of A
example
A
The 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
A
=
the
angle
in
quadrant
Q
with
sine
s
143.1301
=
the
angle
in
quadrant
2
with
sine
3
5
S
=
2
×
s
×
c
−
24
25
=
2
×
3
5
×
(−
4
5
)
K
=
c
2
−
s
2
7
25
=
(−
4
5
)
2
−
(
3
5
)
2
Assumptions
•
sin 2A = 2 sin A cos A and cos 2A = cos²A − sin²A.
•
Find cos A from sin²A + cos²A = 1; its sign comes from the quadrant (negative in II and III).
•
The sine is positive in quadrants I and II and negative in III and IV.
Step-by-step
WE KNOW
s = 3/5, Q = 2
FIND
cos A (c), the angle (A), sin 2A (S), cos 2A (K)
Step 1 · Find cos A (c)
Use c = ±√(1 − s²), the sign from quadrant Q
cos²
A
=
1
−
sin²
A
;
the
root
is
negative
in
quadrants
II
and
III,
positive
in
I
and
IV.
c
=
−
√
1 −
s
2
c
=
−√(1
−
(
3
5
)
2
)
c
=
−
√
1 − 0.36
c
=
−
√
0.64
c
=
−
4
5
Step 2 · Find the angle (A)
Use A = the angle in quadrant Q with sine s
sin⁻¹ gives an angle from −90° to 90°; move it into the quadrant (180° − it in II and III).
A
=
180°
−
sin⁻¹(
s
)
A
=
180°
−
sin⁻¹(
3
5
)
A
=
180°
−
36.8699°
A
=
143.1301°
Step 3 · Find sin 2A (S)
Use S = 2 × s × c
The double-angle formula for sine.
S
=
2
×
3
5
×
(−
4
5
)
S
=
1.2
×
(−
4
5
)
S
=
−
24
25
Step 4 · Find cos 2A (K)
Use K = c² − s²
The double-angle formula for cosine.
K
=
(−
4
5
)
2
−
(
3
5
)
2
K
=
0.64
−
0.36
K
=
7
25
Check
Put the numbers back in. Both sides should match.
−
4
5
=
−
√
1 −
3
5
²
✓
sin(143.1301°)
=
3
5
✓
−
24
25
=
2
×
3
5
×
(−
4
5
)
✓
7
25
=
(−
4
5
)
2
−
(
3
5
)
2
✓
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