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Stitz-Zeager_College_Algebra_e-book

# Weve seen this matrix sin cos most recently in the

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Unformatted text preview: 5π 3 θ= x 5π 4 x Q P √ P has rectangular coordinates (2, −2 3) π P has polar coordinates 4, 53 Q has rectangular coordinates (−3, −3) √π Q has polar coordinates 3 2, 54 3. The point R(0, −3) lies along the negative y -axis. While we could go through the usual computations4 to ﬁnd the polar form of R, in this case we can ﬁnd the polar coordinates of R using the deﬁnition. Since the pole is identiﬁed with the origin, we can easily tell the point R is 3 units from the pole, which means in the polar representation (r, θ) of R we know r = ±3. π Since we require r ≥ 0, we choose r = 3. Concerning θ, the angle θ = 32 satisﬁes 0 ≤ θ < 2π 3π with its terminal side along the negative y -axis, so our answer is 3, 2 . To check, we note π π x = r cos(θ) = 3 cos 32 = (3)(0) = 0 and y = r sin(θ) = 3 sin 32 = 3(−1) = −3. 4. The point S (−3, 4) lies in Quadrant II. With x = −3 and y = 4, we get r2 = (−3)2 +(4)2 = 25 so r = ±5. As usual, we choose r = 5 ≥ 0 and proceed to determine θ. We have tan(θ) = 4 Since x = 0, we would have to determine θ ge...
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