Answers to Questions on Trigonometric Functions


    1. Question: Show that 1 + tan2(x) = sec2(x)

      Answer: tan2(x) + 1 = sin2(x)/cos2(x) + 1 = [sin2(x) + cos2(x)]/cos2(x) = 1/cos2(x) = (1/cos(x))2 = sec2(x). Thus tan2(x) + 1 = sec2(x).

    2. Question: Show that 1 + cot2(x) = csc2(x)

      Answer: Solve similarly to above or:

      Divide both sides of the identity sin2(x) + cos2(x) = 1 by sin2(x) to get
      sin2(x)/sin2(x) + cos2(x)/sin2(x) = 1/sin2(x)
      Thus 1 + cot2(x) = csc2(x).

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  1. Draw graphs of each of the functions. Give the period and any vertical asymptotes. Comment on the amplitudes of these functions.

    You can check these graphs with your graphing calculator. To graph sec (x), enter it as 1/cos (x). Do not use the "cos -1" button (see notes on notation in the Field Guide Lesson.). Similarly, enter 1/sin (x) for csc (x) and 1/tan (x) or cos (x) / sin (x) for cot (x).

    1. Question: cot (x)

      Answer: Cot (x) has period and vertical asymptotes at multiples of (where sin (x) = 0).

    2. Question: sec (x)

      Answer: Sec (x) has period 2 and vertical asymptotes at odd multiples of /2 (where cos (x) = 0).

    3. Question: csc (x)

      Answer: Csc (x) has period 2 and vertical asymptotes at multiples of (where sin (x) = 0).

    These functions do not oscillate up and down within a finite distance from the mid-line y = 0. Thus they do not have a finite amplitude.

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  2. Give the amplitude and period for each of the following functions. Sketch their graphs.

    1. Question: g(x) = 3cos 2x

      Answer: Amplitude = 3, period = .

    2. Question: f(x) = -2.7sin 2(x - /4)

      Answer: Amplitude = 2.7 (negative sign causes a reflection in the x-axis), period = .

    3. Question: g(x) = sin ((1/2)(x + /3)

      Answer: Amplitude = 1, period = 4 .

    4. Question: h(x) = 1.5cos (x - 1)

      Answer: Amplitude = 1.5, period = 2.

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  3. The tidal variation in Desolation Sound on the west coast of Canada is roughly 4 meters. That is, the difference beween water depth at high tide and at low tide is 4 meters, with successive high tides occurring 12.5 hours apart. Suppose that at Refuge Cove in Desolation Sound, the depth of water in meters is given by

    D(t) = D0 + Acos B(t - t0)
    where t is measured in hours from midnight on June 1, 1996.

    1. Question: What does D0 mean in the context of the problem?

      Answer: D0 is the water level midway between high water and low water.

    2. Question: What is the value of A?

      Answer: A = 2. It is the amplitude of oscillation, and is half of the total tidal variation.

    3. Question: What is the value of B?

      Answer: B = (2 )/12.5 = 0.503 (approximately).

    4. Question: Give the physical meaning of t0 for this problem.

      Answer: t0 is the time of the first high tide after midnight on June 1, 1996.

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