Math Problem Statement

Solution

Let's analyze and solve the given problems based on the graph provided and the function f(x)=2cos2x0.5f(x) = 2 \cos 2x - 0.5.

Problem 1(a)

What are the values of the amplitude and period of the function f(x)=2cos2x0.5f(x) = 2 \cos 2x - 0.5?

Amplitude: The amplitude of a cosine function acos(bx+c)+da \cos(bx + c) + d is given by the absolute value of aa. For the given function: f(x)=2cos2x0.5f(x) = 2 \cos 2x - 0.5 The coefficient of the cosine function is 2, so the amplitude is: Amplitude=2=2 units\text{Amplitude} = |2| = 2 \text{ units}

Period: The period of a cosine function acos(bx+c)+da \cos(bx + c) + d is given by: Period=2πb\text{Period} = \frac{2\pi}{|b|} For the given function, b=2b = 2, so the period is: Period=2π2=π units\text{Period} = \frac{2\pi}{2} = \pi \text{ units}

Problem 1(b)

Find the xx-coordinates of the points of intersection for 0x3600^\circ \leq x \leq 360^\circ between f(x)=2cos2x0.5f(x) = 2 \cos 2x - 0.5 and g(x)=sinxg(x) = \sin x.

To find the points of intersection, set f(x)=g(x)f(x) = g(x): 2cos2x0.5=sinx2 \cos 2x - 0.5 = \sin x

This equation can be solved numerically or graphically as it involves both cos2x\cos 2x and sinx\sin x.

  1. Use the double-angle identity for cosine: cos2x=2cos2x1\cos 2x = 2 \cos^2 x - 1 Thus, the equation becomes: 2(2cos2x1)0.5=sinx2 (2 \cos^2 x - 1) - 0.5 = \sin x 4cos2x20.5=sinx4 \cos^2 x - 2 - 0.5 = \sin x 4cos2x2.5=sinx4 \cos^2 x - 2.5 = \sin x

  2. Solve this numerically by finding the values of xx where this equation holds in the interval 0x3600^\circ \leq x \leq 360^\circ.

Since this is a complex equation to solve algebraically, we will consider it graphically or use numerical methods.

From the graph: Let's approximate the xx-coordinates of the intersections. Observing the graph of 2cos2x0.52 \cos 2x - 0.5 and comparing it with sinx\sin x, we can estimate the intersections:

  • At approximately x38.0x \approx 38.0^\circ
  • At approximately x142.0x \approx 142.0^\circ
  • At approximately x218.0x \approx 218.0^\circ
  • At approximately x322.0x \approx 322.0^\circ

Answers: x=38.0,142.0,218.0,322.0x = 38.0^\circ, 142.0^\circ, 218.0^\circ, 322.0^\circ

Summary

  1. Amplitude: 2 units
  2. Period: π\pi units
  3. Intersections: x=38.0,142.0,218.0,322.0x = 38.0^\circ, 142.0^\circ, 218.0^\circ, 322.0^\circ

If you have any questions or need further details, feel free to ask!

Related Questions:

  1. What is the general formula for the amplitude of a trigonometric function?
  2. How do you determine the period of a trigonometric function?
  3. What are the double-angle identities in trigonometry?
  4. How do you solve trigonometric equations graphically?
  5. How can you find the intersection points of two trigonometric functions numerically?
  6. What is the significance of the phase shift in trigonometric functions?
  7. How do trigonometric identities help in solving complex trigonometric equations?
  8. What methods can be used to approximate solutions of trigonometric equations?

Tip:

To solve complex trigonometric equations involving multiple angles and different functions, consider using graphical or numerical methods when algebraic manipulation becomes too cumbersome.

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Math Problem Analysis

Mathematical Concepts

Trigonometry

Formulas

Amplitude of a cosine function
Period of a cosine function
Double-angle identity for cosine

Theorems

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Suitable Grade Level

High School