A sinusoidal function has an amplitude of

Question

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

Answer 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