College Trigonometry, 2011a

730 Foundations of Trigonometry
different than the stated solution of θ = 11π
6
+2πk for integers k, we leave it to the reader to show
they represent the same list of angles.
10.2.1 Beyond the Unit Circle
We began the section with a quest to describe the position of a particle experiencing circular motion.
In defining the cosine and sine functions, we assigned to each angle a position on the Unit Circle. In
this subsection, we broaden our scope to include circles of radius r centered at the origin. Consider
for the moment the acute angle θ drawn below in standard position. Let Q(x, y) bethepointon
theterminalsideofθ which lies on the circle x 2 + y 2 = r 2 ,andletP (x ′ ,y ′ )bethepointonthe
terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and
Q to create two right triangles, ΔOPA and ΔOQB. These triangles are similar, 10 thus it follows
that x x
= r ′ 1 = r, sox = rx′ and, similarly, we find y = ry ′ . Since, by definition, x ′ = cos(θ) and
y ′ =sin(θ), we get the coordinates of Q to be x = r cos(θ) andy = r sin(θ). By reflecting these
points through the x-axis, y-axis and origin, we obtain the result for all non-quadrantal angles θ,
and we leave it to the reader to verify these formulas hold for the quadrantal angles.
y
r
1
θ
1
Q (x, y)
P (x ′ ,y ′ )
r
x
y
1
P (x ′ ,y ′ )
Q(x, y) =(r cos(θ),rsin(θ))
θ
O A(x ′ , 0) B(x, 0)
x
Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of
the circle is r = √ x 2 + y 2 , we can also express cos(θ) andsin(θ) in terms of the coordinates of Q.
These results are summarized in the following theorem.
Theorem 10.3. If Q(x, y) is the point on the terminal side of an angle θ, plotted in standard
position, which lies on the circle x 2 + y 2 = r 2 then x = r cos(θ) andy = r sin(θ). Moreover,
cos(θ) = x r =
x
√ and sin(θ) = y
x 2 + y 2 r = y
√
x 2 + y 2
10 Do you remember why?