Question

How do you write `r=2sec(theta+pi/4)` in rectangular form?

Answer 1

Multiply both sides by the `cos(theta+pi/4)`
Use the identity `cos(a+b)=cos(a)cos(b)-sin(a)sin(b)`
Substitute x for `rcos(theta)` y for `rsin(theta)`

Explanation:

Given: `r = 2sec(theta+pi/4)`

Multiply both sides by `cos(theta+pi/4)`:

`rcos(theta+pi/4) = 2`

Use the identity `cos(a+b)=cos(a)cos(b)-sin(a)sin(b)` to substitute `cos(theta)cos(pi/4) - sin(theta)sin(pi/4)` for `cos(theta+pi/4)`:

`r(cos(pi/4)cos(theta) - sin(pi/4)sin(theta)) = 2`

The sine and cosine are both `sqrt(2)/2` at `pi/4`

`r(sqrt(2)/2cos(theta) - sqrt(2)/2sin(theta)) = 2`

Use the distributive property:

`sqrt(2)/2rcos(theta) - sqrt(2)/2rsin(theta) = 2`

Multiply both sides by `sqrt(2)`

`rcos(theta) - rsin(theta) = 2sqrt2`

Substitute x for `rcos(theta)` and y for `rsin(theta)`

`x - y = 2sqrt2`