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 $T_{f}= \frac{2\pi}{b}$.

f is sine function where f (0)=0, then c=0; with period $\pi$, then $f(\theta)=asin 2\theta$, because $T_{f}= \frac{2 \pi }{2} = \pi$. 

To find a we consider the condition $f( \frac{ \pi }{4} )=4$, from where $asin2* \frac{\pi}{4} =a*sin \frac{ \pi }{2} =a=4$.

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 $g(\theta)=2sin\theta$.