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10 August, 12:18

Express the equation in rectangular form

a) r = 2sin (2Θ)

b) r = 4cos (3Θ)

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Answers (1)
  1. 10 August, 16:13
    0
    a. sqrt (x^2+y^2) * (x^2+y^2) = 4xy

    b. (x^2+y^2) ^2 = 4x{ (x^2 - 3 y^2}

    Step-by-step explanation:

    a) r = 2sin (2Θ)

    r * = 2 sin 2 theta

    We know sin 2theta = 2 sin theta cos theta

    r = 2 * 2 sin theta cos theta

    sin theta = y/r and cos theta = x/r

    r = 4 y/r * x/r

    r = 4xy / r^2

    Multiply each side by r^2

    r^3 = 4xy

    r * r^2 = 4xy

    We know r = sqrt (x^2+y^2) and r^2 = (x^2+y^2)

    sqrt (x^2+y^2) * (x^2+y^2) = 4xy

    b) r = 4cos (3Θ)

    we know that cos 3 theta = cos^3 (theta) - 3 sin^2 (theta) cos (theta)

    r = 4 * cos^3 (theta) - 3 sin^2 (theta) cos (theta)

    Factor out a cos theta

    r = 4 * cos (theta) { cos^2 (theta) - 3 sin^2 (theta) }

    We know that sin theta = y/r and cos theta = x/r

    r = 4 (x/r) { (x/r) ^2 - 3 (y/r) ^2}

    r = 4 * (x/r) { (x^2/r^2 - 3 y^2/r^2}

    Multiply by r

    r^2 = 4x { (x^2/r^2 - 3 y^2/r^2}

    Multiply by r^2

    r^2 * r^2 = 4x { (x^2 - 3 y^2}

    r^4 = 4x{ (x^2 - 3 y^2}

    We know r^2 = (x^2+y^2)

    (x^2+y^2) ^2 = 4x{ (x^2 - 3 y^2}
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