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brilliants [131]
3 years ago
8

Express the equation in rectangular form a) r = 2sin(2Θ) b) r = 4cos(3Θ)

Mathematics
1 answer:
Dafna11 [192]3 years ago
5 0

Answer:

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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