Convert to Rectangular Form...8

[nycmathguy]

Can you get me started with 15 by providing the first 2 steps? I will then try solving the problem on my own. Thank you.

 

[MathLover1]

15.

r*cos(θ-π/6)=2
which is same as r *sin(θ+ π/3)=2

Apply the formula
sin(α+β)=cos(α)sin(β)+sin(α)cos(β) with α=θ and β= π/3
sin(α+β)=cos(θ)sin(π/3)+sin(θ)cos(π/3)
sin(α+β)=sin(θ)/2 + 1/2 sqrt(3) cos(θ)
=>
r( sin(θ)/2+ sqrt(3)cos(θ)/2)=2

From x=r*cos(θ) and y=r*sin(θ), we have that cos(θ)=x/r and sin(θ)=y/r

r( (y/r)/2+ sqrt(3)(x/r)/2)=2

r ((sqrt(3) x)/(2 r) + y/(2 r)) = 2..........simplify, cancel r

(sqrt(3) x + y)/2= 2
sqrt(3) x + y= 4

y=-sqrt(3) x +4

 

[nycmathguy]

15.

r*cos(θ-π/6)=2
which is same as r *sin(θ+ π/3)=2

Apply the formula
sin(α+β)=cos(α)sin(β)+sin(α)cos(β) with α=θ and β= π/3
sin(α+β)=cos(θ)sin(π/3)+sin(θ)cos(π/3)
sin(α+β)=sin(θ)/2 + 1/2 sqrt(3) cos(θ)
=>
r( sin(θ)/2+ sqrt(3)cos(θ)/2)=2

From x=r*cos(θ) and y=r*sin(θ), we have that cos(θ)=x/r and sin(θ)=y/r

r( (y/r)/2+ sqrt(3)(x/r)/2)=2

r ((sqrt(3) x)/(2 r) + y/(2 r)) = 2..........simplify, cancel r

(sqrt(3) x + y)/2= 2
sqrt(3) x + y= 4

y=-sqrt(3) x +4

Thank you. This one is challenging.