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Solutions to the exercises

Exercise 1

y 1 = -1 + 4 sin ( 2 π 4 ( x - 2 ) )
y 2 = 4 sin ( 2 π 20 x )
y 3 = 4 + 2 sin ( 2 π 10 x )
y 4 = 5 + 2 sin ( 2 π 8 ( x + 4 ) )

Exercise 2
a

y 1 = -1 + 4 sin ( 2 π 4 ( x + 2 ) )
y 1 = -1 + 4 cos ( 2 π 4 ( x + 1 ) )
y 1 = -1 + 4 cos ( 2 π 4 ( x - 3 ) )

b

-1 + 4 sin ( 2 π 4 ( x + 2 ) ) = - gives you sin ( π 2 ( x + 2 ) ) = - 1 4 and therefore π 2 ( x + 2 ) -0,253 + k 2 π π 2 ( x + 2 ) 3,394 + k 2 π .
You find: x -2,161 + k 4 x = 0,161 + k 4 .

Exercise 3
a

f ( x ) = 1 + 2 sin ( 2 ( x - 1 6 π ) )

b

f ( 0 ) = 1 - 3

c

f ( x ) = 0 gives sin ( 2 ( x - 1 6 π ) ) = - 1 2 and so 2 ( x - 1 6 π ) = - 1 6 π + k 2 π 2 ( x - 1 6 π ) = 1 1 6 π + k 2 π .
This way you find: x = 1 12 π + k π x = 3 4 π + k π .
The solution to the inequality is - 1 4 π x 1 12 π + k π .

Exercise 4
a

h D = 10 + 8 sin ( 0,6 ) 14,52 m
h C = 10 + 4 sin ( 3,6 ) 8,23 m.

b

h D ( t ) = 10 + 8 cos ( 2 π 8 t )

c

h C ( t ) = 10 + 4 cos ( 1 4 π ( t - 3 ) ) , so h C ( 1413,25 ) 9,22 m.

d

h C = 12 gives cos ( 1 4 π ( t - 3 ) ) = 1 2 and so t = 1 2 3 + k 8 t = 4 1 3 + k 8 .
So every rotation you are 4 1 3 - 1 2 3 = 2 2 3 s above 12 m.

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