How do you write the equation $r = 11 sec (θ + \frac{7 π}{6})$ $r=11sec(theta+(7pi)/6)$ in rectangular form?

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Answer 1 · 2018-03-20T17:59:39

Given: $r = 11 sec (θ + \frac{7 π}{6})$ $r=11sec(theta+(7pi)/6)$

Because we know that $cos (A) sec (A) = 1$ $cos(A)sec(A) = 1$ , we shall multiply both sides by $cos (θ + \frac{7 π}{6})$ $cos(theta+(7pi)/6)$ :

$r cos (θ + \frac{7 π}{6}) = 11$ $rcos(theta+(7pi)/6)=11$

Use the identity $cos (A + B) = cos (A) cos (B) - sin (A) sin (B)$ $cos(A+B) = cos(A)cos(B)-sin(A)sin(B)$

where $A = θ and B = \frac{7 π}{6}$ $A = theta and B = (7pi)/6$

$r (cos (θ) cos (\frac{7 π}{6}) - sin (θ) sin (\frac{7 π}{6})) = 11$ $r(cos(theta)cos((7pi)/6)-sin(theta)sin((7pi)/6))=11$

Distribute the factor, r:

$r cos (θ) cos (\frac{7 π}{6}) - r sin (θ) sin (\frac{7 π}{6}) = 11$ $rcos(theta)cos((7pi)/6)-rsin(theta)sin((7pi)/6)=11$

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

$cos (\frac{7 π}{6}) x - sin (\frac{7 π}{6}) y = 11$ $cos((7pi)/6)x-sin((7pi)/6)y=11$

Calculate the constants:

$- \frac{\sqrt{3}}{2} x + \frac{1}{2} y = 11$ $-sqrt3/2x+1/2y=11$

Multiply both sides by 2:

$- \sqrt{3} x + y = 22$ $-sqrt3x+y=22$

Add $\sqrt{3} x$ $sqrt3x$ to both sides:

$y = \sqrt{3} x + 22$ $y=sqrt3x+22$

How do you write the equation r=11sec(theta+(7pi)/6)r=11sec(θ+7π6)r=11sec(theta+(7pi)/6) in rectangular form?