Find the range and period of the function 𝑓 of 𝜃 equals 14 over five sin of four 𝜃 by drawing the function between zero and two 𝜋.
Let’s just begin by recalling what we mean by the range and the period of the function. Whilst the domain is the set of values we can input into the function, the range is the output. It’s all possible values that 𝑓 of 𝜃 itself can take. We also know that the sine function is periodic. That is, it repeats. And its period is the time taken to complete one full wave.
Now, this question tells us we’re going to need to draw the function between zero and two 𝜋. Before we do though, let’s recall what the graph of 𝑓 of 𝜃 equals sin of 𝜃 looks like. It looks a little something like this. Between zero and two 𝜋, the graph of sin 𝜃 completes one full wave. That means the period of the graph 𝑓 of 𝜃 equal sin 𝜃 is two 𝜋.
Because the graph has rotational symmetry, we can say it intersects the 𝜃-axis at 𝜋 and it has a peak at 𝜋 by two, a maximum of 𝜋 by two, and a local minimum at three 𝜋 by two. These occur when 𝑓 of 𝜃 is equal to one and negative one, respectively. So, the graph of 𝑓 of 𝜃 equals sin 𝜃 has a range from negative one to one, and a period of two 𝜋 radians.
Let’s now recall what we know about transforming graphs. If 𝑦 equals 𝑓 of 𝑥 is some graph, then the graph 𝑦 equals 𝑓 of 𝑎𝑥 where 𝑎 is a constant is an enlargement of the original graph in the 𝑥-direction by a scale factor of one over 𝑎. Similarly, the graph of 𝑦 equals 𝑏 times 𝑓 of 𝑥 is also an enlargement. This time, it’s in the 𝑦-direction. It’s a vertical stretch with a scale factor of 𝑏.
Our graph is sin of four 𝜃 multiplied by 14 over five. This means horizontally in the 𝑥-direction, it’s stretched by a scale factor of one-quarter, but vertically, it’s stretched by a scale factor of fourteen-fifths. This means any maximums will occur when 𝑓 of 𝜃 is equal to fourteen-fifths and minimums will occur when 𝑓 of 𝜃 is negative fourteen-fifths.
A scale factor horizontally of one-quarter means that our original wave will repeat four times in the same interval. It will look a little something like this. This time we can see the range is from negative fourteen-fifths to fourteen-fifths. And the period is a quarter of the original period. It’s 𝜋 by two radians. The range of 𝑓 of 𝜃 equals 14 over five sin of four 𝜃 is the closed interval negative fourteen-fifths to fourteen-fifths, and its period is 𝜋 by two radians.