sin 𝜃 =
=
cos 𝜃 =
=
sin(2𝜃) = 2 sin 𝜃 cos 𝜃 = 2
4
5
3
12
24
=2
=
5
25
25
8 + 15 = 𝑐
64 + 225 = 𝑐
289 = 𝑐
𝑐 = 17
sin 𝑎 =
, cos 𝑎 =
cos(2𝑎) = cos 𝑎 − sin 𝑎
=
15
17
−
8
17
=
225 64
161
−
=
289 289 289
sin(𝜃) =
, cos 𝜃 = −
√
(Negative because we are in Quadrant 2)
√
a. sin(2𝜃) = 2 sin 𝜃 cos 𝜃 = 2
√
b. cos(2𝜃) = cos 𝜃 − sin 𝜃 =
c. tan(2𝜃) =
sin 𝜃 = −
=−
(
)
(
)
√
=
−
√
=
c. tan(2𝜃) =
= , cos 𝜃 =
=
= −
∗−
−
𝜃 − sin 𝜃 =
=2
−
=
=−
−
=
=
=
−
=
24
24 25 24
tan 2𝜃 =
= 25 =
∗
=
7
25 7
7
cos 2𝜃
25
sin 2𝜃
=
=
=
sin 2𝜃 = 2 sin 𝜃 cos 𝜃 = 2
cos 2𝜃 = cos
=2 −
− −
)
√
√
=−
(Negative because we are in Quadrant 4), cos 𝜃 =
b. cos(2𝜃) = cos 𝜃 − sin 𝜃 =
=−
(
=
∗
a. sin(2𝜃) = 2 sin 𝜃 cos 𝜃 = 2 −
sin 𝜃 =
√
=2
=−
−
=
=
sin(𝑎) =
, cos(𝑎) =
, sin(𝐵) =
(Negative because angle B lies in Quadrant 2)
, cos(𝐵) = −
a. cos(𝑎 + 𝐵 ) = cos 𝑎 cos 𝐵 − sin 𝑎 sin 𝐵 =
−
−
=−
−
=−
b. sin(𝑎 + 𝐵 ) = sin 𝑎 cos 𝐵 + cos 𝑎 sin 𝐵 =
−
+
=−
+
=
c. tan(𝑎 + 𝐵 ) =
(
)
(
)
=
=
∗−
=−
cos 𝜃 =
sin =
cos
−
=
=
+
=
−
=
−
=
−
=
=
+
=
+
=
+
=
=
√
=
=
√
√
sin 𝜃 =
= , cos 𝜃 =
sin =
−
cos
=
tan
=
=
−
=
+
√
=
√
=
=
−
+
=
+
√
∗
√
=
=
−
+
=
=
=
2 sin cos = sin 2 ∗ = sin 𝜃 Double Angle Identity
sin 𝜃 =
=
=
=
=
=
√
√
=
=
√
√
sin 𝑎 = −
sin =
cos
, cos 𝑎 =
−
=
=
+
−
=
=
+
−
=
=
+
−
=
=
+
=
=
=
=
=
√
=
√
√
=
√
. BUT because the original angle is
between 270 and 360, the half angle must be between 135 and 180, in Quadrant 2. This means that cos
√
tan
=
=
√
=
√
∗−
√
=−
=−
√
.