How do you write $r = 2 sec (θ + \frac{π}{4})$ $r=2sec(theta+pi/4)$ in rectangular form?

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How do you write $r = 2 sec (θ + \frac{π}{4})$ $r=2sec(theta+pi/4)$ in rectangular form?

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Answer 1 · 2017-01-25T18:54:44

Multiply both sides by the $cos (θ + \frac{π}{4})$ $cos(theta+pi/4)$
Use the identity $cos (a + b) = cos (a) cos (b) - sin (a) sin (b)$ $cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$
Substitute x for $r cos (θ)$ $rcos(theta)$ y for $r sin (θ)$ $rsin(theta)$

Explanation:

Given: $r = 2 sec (θ + \frac{π}{4})$ $r = 2sec(theta+pi/4)$

Multiply both sides by $cos (θ + \frac{π}{4})$ $cos(theta+pi/4)$ :

$r cos (θ + \frac{π}{4}) = 2$ $rcos(theta+pi/4) = 2$

Use the identity $cos (a + b) = cos (a) cos (b) - sin (a) sin (b)$ $cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$ to substitute $cos (θ) cos (\frac{π}{4}) - sin (θ) sin (\frac{π}{4})$ $cos(theta)cos(pi/4) - sin(theta)sin(pi/4)$ for $cos (θ + \frac{π}{4})$ $cos(theta+pi/4)$ :

$r (cos (\frac{π}{4}) cos (θ) - sin (\frac{π}{4}) sin (θ)) = 2$ $r(cos(pi/4)cos(theta) - sin(pi/4)sin(theta)) = 2$

The sine and cosine are both $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ at $\frac{π}{4}$ $pi/4$

$r (\frac{\sqrt{2}}{2} cos (θ) - \frac{\sqrt{2}}{2} sin (θ)) = 2$ $r(sqrt(2)/2cos(theta) - sqrt(2)/2sin(theta)) = 2$

Use the distributive property:

$\frac{\sqrt{2}}{2} r cos (θ) - \frac{\sqrt{2}}{2} r sin (θ) = 2$ $sqrt(2)/2rcos(theta) - sqrt(2)/2rsin(theta) = 2$

Multiply both sides by $\sqrt{2}$ $sqrt(2)$

$r cos (θ) - r sin (θ) = 2 \sqrt{2}$ $rcos(theta) - rsin(theta) = 2sqrt2$

Substitute x for $r cos (θ)$ $rcos(theta)$ and y for $r sin (θ)$ $rsin(theta)$

$x - y = 2 \sqrt{2}$ $x - y = 2sqrt2$

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How do you write $r = 2 sec (θ + \frac{π}{4})$ $r=2sec(theta+pi/4)$ in rectangular form?

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Explanation:

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How do you write $r = 2 sec (θ + \frac{π}{4})$ $r=2sec(theta+pi/4)$ in rectangular form?