Question

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

Answer 1

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`