Question #cbbe5

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Question #cbbe5

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Answer 1 · 2017-07-16T22:08:48

Simplifying it first.

Explanation:

$r = 10 csc (θ + \frac{7 π}{4})$ $r = 10csc(theta + (7pi)/4)$
Multiply by sine.
$r sin (θ + \frac{7 π}{4}) = 10$ $r sin(theta + (7pi)/4) = 10$
Now use the sine addition formula:
$sin (θ + \frac{7 π}{4}) = sin (θ) cos (\frac{7 π}{4}) + cos (θ) sin (\frac{7 π}{4})$ $sin(theta + (7pi)/4) = sin(theta)cos((7pi)/4) + cos(theta)sin((7pi)/4)$
Since $\frac{7 π}{4}$ $(7pi)/4$ is in quadrant IV, sine is negative and cosine is positive, with
$sin (\frac{7 π}{4}) = - \frac{\sqrt{2}}{2}$ $sin((7pi)/4) = -sqrt(2)/2$
and
$cos (\frac{7 π}{4}) = \frac{\sqrt{2}}{2}$ $cos((7pi)/4) = sqrt(2)/2$

Therefore we have...
$r sin (θ + \frac{7 π}{4}) = 10$ $r sin(theta + (7pi)/4) = 10$
$r sin (θ) cos (\frac{7 π}{4}) + r cos (θ) sin (\frac{7 π}{4}) = 10$ $rsin(theta)cos((7pi)/4) + rcos(theta)sin((7pi)/4) = 10$
$- r sin (θ) (\frac{\sqrt{2}}{2}) + r cos (θ) (\frac{\sqrt{2}}{2}) = 10$ $-rsin(theta)(sqrt(2)/2) + rcos(theta)(sqrt(2)/2)= 10$
Multiply both sides by $\sqrt{2}$ $sqrt(2)$ . This has the same effect as dividing by $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ .
$- r sin (θ) + r cos (θ) = 10 \sqrt{2}$ $-rsin(theta) + rcos(theta)= 10sqrt(2)$
Now use the standard polar substitutions:
$x = r cos (θ) and y = r sin (θ)$ $x = rcos(theta) and y = rsin(theta)$ .
We have...
$- y + x = 10 \sqrt{2}$ $-y + x = 10sqrt(2)$
Solve for y, or put it in whatever form you desire. This is a line.

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Question #cbbe5

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