Question

How do you convert `(theta)=-7pi/4` to rectangular form?

Answer 1

Equation in rectangular coordinates is `x+y=0`

Explanation:

To convert equation `theta=−(7pi)/4`, remember relation between polar and rectangular coordinates given by

`x=rcostheta` and `y=rsintheta` i.e. `r^2=x^2+y^2` and `theta=tan^(-1)(y/x)` or `tantheta=y/x`

Hence the equation `theta=−(7pi)/4` means

`tan (-(7pi)/4)=y/x`, but `tan(-(7pi)/4)=-1`

Hence equation in rectangular coordinates is `y/x=-1` i.e.

`x+y=0`

Answer 2

The radial line (half line ) y = x, in the first quadrant..

Explanation:

In polar coordinates, `theta` = constant represents a radial line and is a half-line.
The other half in the opposite direction is governed by `theta` = the constant + `pi`.
Indeed, the given equation can be given directly as `theta` = `pi`/4.

Answer 3

one more Explanation

Explanation:

`y=rsin(-7pi/4)`
`=>y=-rsin(7pi/4)` [since `sin(-theta)=-sintheta`]
`=>y=-rsin(2pi-pi/4)`
`=>y=rsin(pi/4)` [in 4th quadrant sin is negative]
`=>y=r/sqrt2`
again
`x=rcos(-7pi/4)` [since `cos(-theta)=costheta`]
`=>x=rcos(7pi/4)`
`=>x=rcos(2pi-pi/4)`
`=>x=rcos(pi/4)` [in 4th quadrant cos is positive]
`=>x=r/sqrt2`
hence
`x=y`