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Using the graphs of trigonometric functions to solve real-world problems

Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function .

Using trigonometric functions to solve real-world scenarios

Suppose the function y = 5 tan ( π 4 t ) marks the distance in the movement of a light beam from the top of a police car across a wall where t is the time in seconds and y is the distance in feet from a point on the wall directly across from the police car.

  1. Find and interpret the stretching factor and period.
  2. Graph on the interval [ 0 , 5 ] .
  3. Evaluate f ( 1 ) and discuss the function’s value at that input.
  1. We know from the general form of y = A tan ( B t ) that | A | is the stretching factor and π B is the period.
    A graph showing that variable A is the coefficient of the tangent function and variable B is the coefficient of x, which is within that tangent function.

    We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period.

    The period is π π 4 = π 1 4 π = 4. This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.

  2. To graph the function, we draw an asymptote at t = 2 and use the stretching factor and period. See [link]
    A graph of one period of a modified tangent function, with a vertical asymptote at x=4.
  3. period: f ( 1 ) = 5 tan ( π 4 ( 1 ) ) = 5 ( 1 ) = 5 ; after 1 second, the beam of has moved 5 ft from the spot across from the police car.
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Access these online resources for additional instruction and practice with graphs of other trigonometric functions.

Key equations

Shifted, compressed, and/or stretched tangent function y = A tan ( B x C ) + D
Shifted, compressed, and/or stretched secant function y = A sec ( B x C ) + D
Shifted, compressed, and/or stretched cosecant function y = A csc ( B x C ) + D
Shifted, compressed, and/or stretched cotangent function y = A cot ( B x C ) + D

Key concepts

  • The tangent function has period π .
  • f ( x ) = A tan ( B x C ) + D is a tangent with vertical and/or horizontal stretch/compression and shift. See [link] , [link] , and [link] .
  • The secant and cosecant are both periodic functions with a period of 2 π . f ( x ) = A sec ( B x C ) + D gives a shifted, compressed, and/or stretched secant function graph. See [link] and [link] .
  • f ( x ) = A csc ( B x C ) + D gives a shifted, compressed, and/or stretched cosecant function graph. See [link] and [link] .
  • The cotangent function has period π and vertical asymptotes at 0 , ± π , ± 2 π , ... .
  • The range of cotangent is ( , ) , and the function is decreasing at each point in its range.
  • The cotangent is zero at ± π 2 , ± 3 π 2 , ... .
  • f ( x ) = A cot ( B x C ) + D is a cotangent with vertical and/or horizontal stretch/compression and shift. See [link] and [link] .
  • Real-world scenarios can be solved using graphs of trigonometric functions. See [link] .

Section exercises

Verbal

Explain how the graph of the sine function can be used to graph y = csc x .

Since y = csc x is the reciprocal function of y = sin x , you can plot the reciprocal of the coordinates on the graph of y = sin x to obtain the y -coordinates of y = csc x . The x -intercepts of the graph y = sin x are the vertical asymptotes for the graph of y = csc x .

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Questions & Answers

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If auger is pair are the roots of equation x2+5x-3=0
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Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
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Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
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Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
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-42m²+60m-18
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-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
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Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
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12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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