Question

The functions f (theta) and g (theta) are sine functions, where f (0) equals g (0) equals 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.

Answer 1

If period of
`f(\theta)` is one-half the period of
`g(\theta)` and

`g(\theta)` has a period of 2π, then
`T_(g) =2T_(f)=2 \pi` and
`T_(f)= \pi`.

To find the period of sine function
`f(\theta)=asin(b\theta+c)` we use the rule
`T_(f)= (2\pi)/(b)`.

f is sine function where f (0)=0, then c=0; with period
`\pi`, then
`f(\theta)=asin 2\theta`, because
`T_(f)= (2 \pi )/(2) = \pi`.

To find a we consider the condition
`f( ( \pi )/(4) )=4`, from where
`asin2* (\pi)/(4) =a*sin ( \pi )/(2) =a=4`.

If the amplitude of
`f(\theta)` is twice the amplitude of
`g(\theta)` , then
`g(\theta)` has a product factor twice smaller than
`f(\theta)` and while period of
`g(\theta)` is 2π and g(0)=0, we can write
`g(\theta)=2sin\theta`.