Quiz Discussion: Use The Given Information To Find The E
Quiz Discussionuse The Given Information About To Find The Exact Val
Use the given information about angles and functions to find the exact values of specified trigonometric functions and solve related problems. Tasks include analyzing and graphing functions, converting between coordinate systems, and applying vector operations and conic section classifications. Draw diagrams where appropriate to illustrate the problems, and clearly label all known sides, angles, directions, and coordinate points. Utilize the Law of Cosines or Law of Sines to solve triangles, and convert between polar and Cartesian coordinates as needed. For vector problems, compute magnitudes, vector sums, differences, and angles between vectors, and represent vectors in component form.
Paper For Above instruction
Understanding and applying fundamental trigonometric concepts is essential for solving a broad range of mathematical problems, including those involving angles, functions, vectors, and conic sections. This paper explores various mathematical tasks based on the provided instructions, emphasizing the importance of exact calculations, graphing, and coordinate transformations.
Exact Values of Trigonometric Functions
Given an angle θ where cos θ = 3/5 and 0
cos(θ/2) = ±√[(1 + cos θ)/2].
Since θ is in the first quadrant, cos(θ/2) is positive. Substituting cos θ = 3/5 yields:
cos(θ/2) = √[(1 + 3/5)/2] = √[(8/5)/2] = √[(8/5) * (1/2)] = √(8/10) = √(4/5) = 2/√5 = (2√5)/5.
Thus, the exact value of cos(θ/2) is (2√5)/5.
Graphing and Analyzing Trigonometric Functions
Consider the function y = cos(1/2 x + π/2) + 3. To analyze this function, we determine its amplitude, period, and phase shift:
- Amplitude: The coefficient before the cosine function is 1, so the amplitude is 1.
- Period: The period of a cosine function y = cos(kx + c) is 2π / |k|. Here, k = 1/2, so:
Period = 2π / (1/2) = 2π * 2 = 4π.
- Phase shift: The phase shift is given by -c / k, where c = π/2. Thus:
Phase shift = - (π/2) / (1/2) = -π/2 * 2 = -π.
The graph of one period of this function starts at x = -π, repeats every 4π units, and has an amplitude of 1 with a vertical shift of +3, thus oscillating between y = 2 and y = 4.
Graphing this, you would plot key points at x = -π, -π + 2π, -π + 4π, noting that at critical points, the cosine reaches its maximum and minimum, then sketch the wave accordingly.
Solving Trigonometric Equations
Next, consider the equation sin(x) + cos(x) = 1 in the interval [0, 2π]. Using identities, we can rewrite the sum as:
sin(x) + cos(x) = √2 sin(x + π/4).
Set this equal to 1:
√2 sin(x + π/4) = 1 ⇒ sin(x + π/4) = 1/√2 = √2/2.
The solutions for sin θ = √2/2 in [0, 2π] are θ = π/4 and 3π/4. Therefore,
x + π/4 = π/4, 3π/4, which yields:
x = 0, π/2.
Hence, the solutions are x = 0 and x = π/2 in [0, 2π].
Applying the Law of Cosines and Sines in Triangle Problems
Suppose a vehicle travels due west for 30 miles and then changes direction to S68°W for another 30 miles. To determine its distance from the starting point, construct a coordinate diagram. The initial leg along the west direction corresponds to moving along the negative x-axis from (0,0) to (-30, 0). The second movement at an angle of 68° south of west positions the vehicle in the coordinate plane. Using the Law of Cosines, we can compute the resultant displacement:
Let the sides be 30 miles each, with an included angle of 68° between two vectors pointing west and S68°W. The resultant is given by:
r = √(30^2 + 30^2 - 23030*cos 68°) = √(900 + 900 - 1800 cos 68°).
Calculating cos 68° ≈ 0.3746,
r ≈ √(1800 - 1800 * 0.3746) ≈ √(1800 - 674.28) ≈ √1125.72 ≈ 33.56 miles.
The vehicle is approximately 33.56 miles from the start, measured straight-line distance.
Polar Coordinates and Conversion to Cartesian Coordinates
Given polar coordinates (r, θ), the Cartesian equivalents are obtained through:
- x = r cos θ
- y = r sin θ
For example, the point (8, 3π/4) converts to:
x = 8 cos (3π/4) = 8 * (-√2/2) = -4√2
y = 8 sin (3π/4) = 8 * (√2/2) = 4√2
Resulting in Cartesian coordinates (-4√2, 4√2). Similarly, for the Cartesian point (-6, 0), the polar coordinates are:
r = √((-6)^2 + 0^2) = 6, and θ = π (since pointing directly left), thus (6, π).
These conversions facilitate plotting and analyzing points across coordinate systems.
Polar Function Graphing and Solving
For the function r = 2 - 4 cos(θ/2), specific points can be identified by substitution of θ-values. Given θ = -π/3:
r = 2 - 4 cos(-π/6) = 2 - 4 * (√3/2) = 2 - 2√3.
At θ = 0:
r = 2 - 4 cos 0 = 2 - 4 * 1 = -2.
Setting r = 0 to find corresponding θ:
0 = 2 - 4 cos(θ/2) ⇒ 4 cos(θ/2) = 2 ⇒ cos(θ/2) = 1/2, thus θ/2 = π/3 or 5π/3, leading to θ = 2π/3 or 10π/3. The \(\theta\) values correspond to angles where the graph intersects the pole, indicating multiple solutions depending on the range.
Graphing this function involves plotting points at these specific θ-values and sketching the shape accordingly, noting its lemniscate or limacon features, which are typical in such polar equations.
Vector Operations in Standard Form
Given vectors u = and v = , the magnitude of u is
||u|| = √(4^2 + 5^2) = √(16 + 25) = √41 ≈ 6.40.
The difference vector u - v is:
= .
Its magnitude:
||u - v|| = √(5^2 + 20^2) = √(25 + 400) = √425 ≈ 20.62.
The angle θ between the vectors can be found using the dot product:
u · v = (4)(-1) + (5)(-15) = -4 - 75 = -79.
θ = arccos[(u · v) / (||u|| * ||v||)].
First, find ||v||:
||v|| = √((-1)^2 + (-15)^2) = √(1 + 225) = √226 ≈ 15.03.
Therefore,
θ = arccos(-79 / (6.40 * 15.03)) ≈ arccos(-79 / 96.11) ≈ arccos(-0.822).
Hence, θ ≈ 145.8°, indicating the vectors form an obtuse angle.
Classification of Quadratic Equations and Conic Sections
The given equation (y + 2)^2 = 4(x + 6) represents a parabola. Its vertex is at (-6, -2). The focus and directrix can be calculated by recognizing the standard form of a parabola opening rightward:
Standard form: (y - k)^2 = 4p(x - h).
Here, p = 1, so the focus is at:
(h + p, k) = (-6 + 1, -2) = (-5, -2).
The directrix is the vertical line x = h - p = -6 - 1 = -7. The parabola opens to the right, with the vertex at (-6, -2).
If the equation were an ellipse, hyperbola, or circle, similar procedures involving completing the square, examining coefficients, and assessing standard forms would be used to classify and identify center points, vertices, foci, and axes.
Vectors: Dot Product and Angle Calculation
Given the vectors \(\vec{u} = \) and \(\vec{v} = \), the dot product is:
\(\vec{u} \cdot \vec{v} = (-2)(3) + (-8)(2) = -6 - 16 = -22.\)
The magnitude of \(\vec{u}\) is:
\(\sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68} \approx 8.25,\)
and for \(\vec{v}\):
\(\sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.61.\)
The angle \(\theta\) between the vectors is calculated via:
\(\theta = \arccos\left(\frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| \times ||\vec{v}||}\right) = \arccos(-22 / (8.25 \times 3.61)) \approx \arccos(-22 / 29.78) \approx \arccos(-0.739).\)
Thus, \(\theta \approx 137.1^\circ\).
Conclusion
Mastering these foundational concepts in trigonometry, coordinate transformation, vector algebra, and conic sections enables effective problem-solving across a variety of mathematical contexts. Precise calculations, careful diagramming, and systematic application of identities and formulas are key to deriving exact solutions and visual understanding of geometric relationships. Proficiency in converting between coordinate systems and analyzing functions graphically enhances overall mathematical literacy, preparing learners for advanced topics in calculus, physics, and engineering.
References
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- Larson, R., Hostetler, R. P., & Edwards, B. H. (2015). Calculus (10th ed.). Cengage Learning.
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- Stewart, J. (2015). Precalculus: Mathematics for Calculus (7th ed.). Cengage Learning.