Math Problem Statement

Solve the mathematical problems provided in the uploaded image, including simplifying trigonometric expressions, solving systems of equations, and converting between coordinate systems.

Solution

Here’s a detailed explanation and solution to each of the problems in the image:


6. Simplify: cot(sin⁻¹(2/3))

  1. Let θ=sin1(2/3)\theta = \sin^{-1}(2/3). Then sin(θ)=2/3\sin(\theta) = 2/3.
  2. Use the Pythagorean identity: cos(θ)=1sin2(θ)=1(2/3)2=14/9=5/9=5/3\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - (2/3)^2} = \sqrt{1 - 4/9} = \sqrt{5/9} = \sqrt{5}/3.
  3. Cotangent is given by: cot(θ)=cos(θ)sin(θ)=5/32/3=52\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{\sqrt{5}/3}{2/3} = \frac{\sqrt{5}}{2}.

Answer: (a) 52\frac{\sqrt{5}}{2}


7. Solve the system of equations: 2x + 5y = 8, 6x + 15y = 24

  1. Rewrite the second equation: 6x+15y=243(2x+5y)=246x + 15y = 24 \rightarrow 3(2x + 5y) = 24, which simplifies to 2x+5y=82x + 5y = 8, identical to the first equation.
  2. Since the two equations are identical, the system has infinitely many solutions.

Answer: (b) Infinitely many solutions


8. Convert rectangular coordinates (1, -1) into polar coordinates

  1. Polar coordinates (r,θ)(r, \theta) are given by:
    • r=x2+y2=12+(1)2=2r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2},
    • θ=tan1(y/x)=tan1(1/1)=tan1(1)\theta = \tan^{-1}(y/x) = \tan^{-1}(-1/1) = \tan^{-1}(-1).
  2. Since the point is in Quadrant IV, θ=7π4\theta = \frac{7\pi}{4} (adjusted for quadrant).

Answer: (b) (2,7π4)(- \sqrt{2}, \frac{7\pi}{4})


9. Find the rectangular coordinates for the point whose polar coordinates are (5, 5π5\pi)

  1. Rectangular coordinates are given by:
    • x=rcos(θ)=5cos(5π)x = r \cos(\theta) = 5 \cos(5\pi),
    • y=rsin(θ)=5sin(5π)y = r \sin(\theta) = 5 \sin(5\pi).
  2. Since cos(5π)=1\cos(5\pi) = -1 and sin(5π)=0\sin(5\pi) = 0,
    • x=5(1)=5x = 5(-1) = -5,
    • y=0y = 0.

Answer: (e) (5,0)(-5, 0)


10. Which point is NOT the same as the polar coordinates (3,5π3)(-3, \frac{5\pi}{3})?

  1. (3,5π3)(-3, \frac{5\pi}{3}) is equivalent to:
    • (3,5π32π)=(3,π3)(-3, \frac{5\pi}{3} - 2\pi) = (-3, -\frac{\pi}{3}),
    • Rotating by multiples of 2π2\pi, the equivalent points are:
      • (3,2π3)(3, \frac{2\pi}{3}),
      • (3,5π3)(3, -\frac{5\pi}{3}),
      • (3,4π3)(3, -\frac{4\pi}{3}),
      • (3,π3)(-3, -\frac{\pi}{3}).
  2. The point (3,5π3)(3, -\frac{5\pi}{3}) is NOT the same.

Answer: (b) (3,5π3)(3, -\frac{5\pi}{3})


If you need further clarification, let me know!


5 Follow-up Questions

  1. Can you generalize how to convert polar coordinates to rectangular coordinates?
  2. How would you solve a system of equations graphically?
  3. Can you derive all equivalent polar coordinates for a given point?
  4. What is the significance of cotangent in trigonometric simplifications?
  5. How would you explain the relationship between rectangular and polar forms to a

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Algebra
Polar and Rectangular Coordinates
Systems of Linear Equations

Formulas

Pythagorean identity: sin^2(θ) + cos^2(θ) = 1
cotangent: cot(θ) = cos(θ)/sin(θ)
Conversion between polar and rectangular coordinates: r = √(x^2 + y^2), θ = tan^(-1)(y/x)

Theorems

Trigonometric identities
Methods for solving systems of linear equations

Suitable Grade Level

Grades 10-12