How do you graph $r = - 2 sin θ$ $r=-2sintheta$ ?

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How do you graph $r = - 2 sin θ$ $r=-2sintheta$ ?

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Answer 1 · 2016-10-14T19:02:10

The way that you graph this is, set your compass to a radius of 1, put the center point at $(0, - 1)$ $(0, -1)$ , and draw a circle.

Explanation:

Multiply both sides by r:

$r^{2} = - 2 r sin (θ)$ $r^2 = -2rsin(theta)$

Substitute $x^{2} + y^{2}$ $x^2 + y^2$ for $r^{2}$ $r^2$ and $y$ $y$ for $r sin (θ)$ $rsin(theta)$

$x^{2} + y^{2} = - 2 y$ $x^2 + y^2 = -2y$

Write $x^{2} a s {(x - 0)}^{2}$ $x^2 as (x - 0)^2$ :

${(x - 0)}^{2} + y^{2} = - 2 y$ $(x - 0)^2 + y^2 = -2y$

Add $2 y + k^{2}$ $2y + k^2$ to both sides:

${(x - 0)}^{2} + y^{2} + 2 y + k^{2} = k^{2}$ $(x - 0)^2 + y^2 + 2y + k^2= k^2$

Using the pattern ${(y - k)}^{2} = x^{2} - 2 k y + k^{2}$ $(y - k)^2 = x^2 - 2ky + k^2$ we observe that we can use the 2y term to find the value of $k and k^{2}$ $k and k^2$ :

$- 2 k y = 2 y$ $-2ky = 2y$

$k = - 1 and k^{2} = 1$ $k = -1 and k^2 = 1$

Substitute this into the equation:

${(x - 0)}^{2} + y^{2} + 2 y + 1 = 1$ $(x - 0)^2 + y^2 + 2y + 1= 1$

This means y terms on the left side can be written as ${(y - - 1)}^{2}$ $(y - -1)^2$ :

${(x - 0)}^{2} + {(y - - 1)}^{2} = 1^{2}$ $(x - 0)^2 + (y - -1)^2 = 1^2$

The way that you graph this is, set your compass to a radius of 1, put the center point at $(0, - 1)$ $(0, -1)$ , and draw a circle.

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How do you graph $r = - 2 sin θ$ $r=-2sintheta$ ?

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How do you graph r=−2sinθr=-2sintheta?

1 Answer

Explanation:

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How do you graph $r = - 2 sin θ$ $r=-2sintheta$ ?