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Math, Trigonometric Functions

A sinusoidal function has an amplitude of
4 units, a period of 90°, and a maximum at
(0, 2).
a) Represent the function with an equation
using a cosine function.
b) Represent the function with an equation
using a sine function.
How would I get the d and c values?
(y=asin(k(x-d))+c, y=acos(k(x-d))+c

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Evan

  1. a = 4
    k = 360°/90° = 4

    so far we would have
    y = 4cos (4x), where x would have to be in degrees
    our first max would be (0,4) but we want it to be (0,2) , so we have drop our
    curve 2 units.
    y = 4cos(4x) - 2
    this meets all your conditions so don't need a horizontal phase shift, the d

    a) y = 4cos (4x) - 2

    b) same a, and k
    y = 4 sin (4x) -2 , this has a max at (22.5°,2) , but we need it at (0,2)
    so we have to shift our curve to the left by 22.5°

    y = 4sin (4(x + 22.5°)) - 2

    formal way:
    we need (0,2) to be on the curve y = 4sin(4(x - d)) - 2
    2 = 4sin(4(0-d) - 2
    4 = 4sin(-4d)
    1 = sin(-4d)
    but I know that sin 90° = 1, or take sineinverse(1) to get 90
    so -4d = 90
    d = -22.5
    making our equation:
    y = 4 sin (4(x+22.5°) - 2

    normally these type of equations would be expressed in radians,
    yours would be
    a) y = 4cos(4 x) - 2 , to get the k value: 2π/k = π/2 ---> k = 4
    b) y = 4sin (4(x + π/8)) - 2

    www.wolframalpha.com/input/?i=y+%3D+4cos%284+x%29+-+2+%2C+y+%3D+4sin+%284%28x+%2B+%CF%80%2F8%29%29+-+2

    notice the two graphs coincide, and all your stated conditions are met

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    mathhelper

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