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

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Answer 1 · 2016-11-19T17:34:48

The equation is $x - y = 4 \sqrt{2}$ $x-y=4sqrt2$

We use,

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

$cos (θ + \frac{π}{4}) = cos θ cos (\frac{π}{4}) - sin θ sin (\frac{π}{4})$ $cos(theta+pi/4)=costhetacos(pi/4)-sinthetasin(pi/4)$

$cos (\frac{π}{4}) = sin (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $cos(pi/4)=sin(pi/4)=sqrt2/2$

The conversion from polar coordnates to cartesian coordinates is made with the following equations

$x = r cos θ$ $x=rcostheta$

and $y = r sin θ$ $y=rsintheta$

We can transform the above equation,

$4 = r cos (θ + \frac{π}{4}) = r cos θ cos (\frac{π}{4}) - r sin θ sin (\frac{π}{4})$ $4=rcos(theta+pi/4)=rcosthetacos(pi/4)-rsinthetasin(pi/4)$

$4 = x \frac{\sqrt{2}}{2} - y \frac{\sqrt{2}}{2}$ $4=xsqrt2/2-ysqrt2/2$

$4 \sqrt{2} = x - y$ $4sqrt2=x-y$

How do you write the equation 4=rcos(theta+pi/4)4=rcos(θ+π4)4=rcos(theta+pi/4) in rectangular form?