Unformatted text preview:
y
that tan(θ) = x = −22 3 = − 3. This tells us θ has a reference angle of π , and since P
3
lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have 0 ≤ θ < 2π ,
π
π
π
so we choose θ = 53 . Hence, our answer is 4, 53 . To check, we convert (r, θ) = 4, 53
1
5π
back to rectangular coordinates and we ﬁnd x = r cos(θ) = 4 cos 3 = 4 2 = 2 and
√
√
π
y = r sin(θ) = 4 sin 53 = 4 − 23 = −2 3, as required. 2. The point √(−3, −3) lies in Quadrant III. Using x = y = −3, we get r2 = (−3)2√ (−3)2 = 18
Q
+
√
so r = ± 18 = ±3 2. Since we are asked for r ≥ 0, we choose r = 3 2. We ﬁnd
−3
tan(θ) = −3 = 1, which means θ has a reference angle of π . Since Q lies in Quadrant III,
4
π
we choose θ = 54 , which satisﬁes the requirement that 0 ≤ θ < 2π . Our ﬁnal answer is
√
√π
√
√
π
(r, θ) = 3 2, 54 . To check, we ﬁnd x = r cos(θ) = (3 2) cos 54 = (3 2) − 22 = −3
√
√
√
π
and y = r sin(θ) = (3 2) sin 54 = (3 2) − 22 = −3, so we are done.
y y θ=...
View
Full Document
 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

Click to edit the document details