Answer:
The points are (3π/4, [tex]\frac{2 + \sqrt{2} }{2}[/tex]) and (7π/4, [tex]\frac{2 - \sqrt{2} }{2}[/tex])
Step-by-step explanation:
Since the two equations r = 1 + cos(θ) and r = 1 − sin(θ) intersect, then
1 + cos(θ) = 1 − sin(θ)
collecting like terms
cos(θ) + sin(θ) = 1 - 1
cos(θ) + sin(θ) = 0
dividing through by cos(θ), we have
cos(θ)/cos(θ) + sin(θ)/cos(θ) = 0/cos(θ)
1 + tan(θ) = 0
tan(θ) = -1
Since tanθ is negative in both the second and fourth quadrant, we have
tan(π - θ) = 1 or tan(2π - θ) = 1
(π - θ) = tan⁻¹1 or (2π - θ) = tan⁻¹1
(π - θ) = π/4 or (2π - θ) = π/4
θ = π - π/4 = 3π/4 or θ = 2π - π/4 = 7π/4
Substituting these values into r = 1 + cos(θ), we have
r = 1 - cos(3π/4) or r = 1 - cos(7π/4)
r = 1 - (-1/√2) or r = 1 - 1/√2
r = 1 + √2/2 or r = 1 - √2/2
r = [tex]\frac{2 + \sqrt{2} }{2}[/tex] or r = [tex]\frac{2 - \sqrt{2} }{2}[/tex]
So, the points are (3π/4, [tex]\frac{2 + \sqrt{2} }{2}[/tex]) and (7π/4, [tex]\frac{2 - \sqrt{2} }{2}[/tex])
There are two possible solutions between [tex]0[/tex] and [tex]2\pi[/tex]:
[tex]\theta_{1} = \frac{3\pi}{4}[/tex], [tex]\theta_{2} = \frac{7\pi}{4}[/tex]
Let be [tex]r_{1} = 1 + \cos \theta[/tex] and [tex]r_{2} = 1 - \sin \theta[/tex], all points of intersection exists when [tex]r_{1} = r_{2}[/tex] for a given [tex]\theta[/tex], that is to say:
[tex]1 + \cos \theta = 1 - \sin \theta[/tex]
[tex]\tan \theta = -1[/tex]
[tex]\theta = \tan^{-1} (-1)[/tex]
As of tangent function is a trigonometric function with a periodicity of [tex]\pi[/tex]. Hence, the set of solutions for the function is:
[tex]\theta = \frac{3\pi}{4} + \pi\cdot i[/tex], for [tex]i \in \mathbb{Z}[/tex]
There are two possible solutions between [tex]0[/tex] and [tex]2\pi[/tex]:
[tex]\theta_{1} = \frac{3\pi}{4}[/tex], [tex]\theta_{2} = \frac{7\pi}{4}[/tex]
We kindly invite to check this question on polar equations: https://brainly.com/question/1269731
