Question

Periodic function help?
The function f(theta) and g(theta) re sine functions, where f(0) = g(0) = 0. The amplitude of f(theta) is twice the amplitude of g(theta). The period of f(theta) is one-half the period of g(theta). If g(theta) has a period of 2pi and f(pi/4) = 4, write the function rule for g(theta). Explain your reasoning.

Ok, so I know the standard form is f(x) = a sin b theta (where a is amplitude and b is the period), but i'm lost on how to use this, but I know it has to have something to do with this.

Answer 1

@amistre64 Can you help?

Answer 2

@Hero Do you know how to solve this?

Answer 3

if g(Θ) has a period of 2pi, then it will have the form
g(Θ) = a sin(Θ)

Answer 4

f(Θ) has period one half of g, so the period of f is π

Answer 5

g(Θ) = a sin(Θ)
f(Θ) = 2a*sin(2Θ)

and you are given that
f(π/4) = 4

Answer 6

Therefore
4 = 2*a sin(π/4)

can you solve it now?

Answer 7

So 4 = sqrt(2)a?

Answer 8

g(Θ) = a sin(2pi theta). I just need the amplitude :/

Answer 9

right, so you can solve for `a`

Answer 10

I get sqrt[8] = a?

4 = sqrt(2)a
16 = 2(a)^2
8 = a^2
sqrt[8] = a?

Answer 11

@perl ?

Answer 12

yes that's `a`

Answer 13

which is simplified as 2√2

Answer 14

Awesome! So g(theta) = 2(sqrt[2]) * (sin(2pi theta)) Thanks!

Answer 15

no thats incorrect

Answer 16

y = sin Θ , has a period of 2pi, you don't want to introduce another factor of 2pi in there

Answer 17

Ooh, so just g(theta) = 2(sqrt[2]) * (sin theta)?

Answer 18

g(Θ) = 2√2 sin(Θ)
f(Θ) = 2(2√2) sin(2Θ)

Answer 19

Awesome. Thanks! This can get a bit confusing for me, so I appreciate the patience!