Stitz-Zeager_College_Algebra_e-book

But we dont so we wont 5 in this solution we take the

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Unformatted text preview: ±3 has an equivalent11 representation which satisfies r = −3. √ π π (b) We take the tangent of both sides the equation θ = 43 to get tan(θ) = tan 43 = 3. √ √ y y Since tan(θ) = x , we get x = 3 or y = x 3. Of course, we pause a moment to wonder √ π if, geometrically, the equations θ = 43 and y = x 3 generate the same set of points.12 The same argument presented in number 1b applies equally well here so we are done. (c) Once again, we need to manipulate r = 1 − cos(θ) a bit before using the conversion formulas given in Theorem 11.7. We could square both sides of this equation like we did in part 2a above to obtain an r2 on the left hand side, but that does nothing helpful for the right hand side. Instead, we multiply both sides by r to obtain r2 = r − r cos(θ). We now have an r2 and an r cos(θ) in the equation, which we can easily handle, but we also have another r to deal with. Rewriting the equation as r = r2 + r cos(θ) 2 and squaring both sides yields r2 = r2 + r cos(θ) . Substituting r2 = x2 + y 2 and 2 r cos(θ) = x gives x2 +...
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