How do you graph the polar equation $3 = r cos (θ + 90^{\circ})$ $3=rcos(theta+90^circ)$ ?

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Answer 1 · 2016-11-15T23:15:04

Draw a horizontal line that crosses the y-axis at -3.

Use the sum of two angles identity:

$cos (a + b) = cos (a) cos (b) - sin (a) sin (b)$ $cos(a + b) = cos(a)cos(b) - sin(a)sin(b)$

To discover that:

$cos (θ + 90^{\circ}) = - sin (θ)$ $cos(theta + 90^@) = -sin(theta)$

Make that substitution:

$3 = - r sin (θ)$ $3 = -rsin(theta)$

Flip and multiply by -1

$r sin (θ) = - 3$ $rsin(theta) = -3$

Substitute y for $r sin (θ)$ $rsin(theta)$ :

$y = - 3$ $y = -3$

Draw a horizontal line that crosses the y-axis at -3.

How do you graph the polar equation 3=rcos(θ+90∘)3=rcos(theta+90^circ)?