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

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Answer 1 · 2018-03-20T17:59:39

Given: #r=11sec(theta+(7pi)/6)#

Because we know that #cos(A)sec(A) = 1#, we shall multiply both sides by #cos(theta+(7pi)/6)#:

#rcos(theta+(7pi)/6)=11#

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

where #A = theta and B = (7pi)/6#

#r(cos(theta)cos((7pi)/6)-sin(theta)sin((7pi)/6))=11#

Distribute the factor, r:

#rcos(theta)cos((7pi)/6)-rsin(theta)sin((7pi)/6)=11#

Substitute #rcos(theta) = x and rsin(theta)=y#

#cos((7pi)/6)x-sin((7pi)/6)y=11#

Calculate the constants:

#-sqrt3/2x+1/2y=11#

Multiply both sides by 2:

#-sqrt3x+y=22#

Add #sqrt3x# to both sides:

#y=sqrt3x+22#