- How do trigonometric functions interact with other functions?
- How do we find zeros of trigonometric functions?
- What does average rate of change look like with trigonometric functions?
Trig Function Compositions
Trigonometric functions can be composed with any of the types of functions that we have already seen. Just as with other function compostions, we need to be mindful of the domains and ranges of our functions.
Find the function below and state its domain and range.
- (a)
- (b)
- (c)
The domain for both and is . The range for is and the range for is . Now we can look at the compositions.
- (a)
The domain of consists of all inputs to whose corresponding outputs are in the domain of . Since the domain of is , all outputs of are in the domain of , so the domain of is the entire domain of , namely .Finding the range of is a bit trickier. Since the range of is only non-negative numbers, only non-negative numbers will be plugged into when evaluating the composition. The question therefore becomes: what outputs of correspond to non-negative inputs? The answer is: all of them! By looking at the graph of the sine function, we can see that all -values have a corresponding non-negative -value. Therefore, the range of , namely , is the range of .
- (b)
- . Remember that means .
All real numbers can be plugged into , and the results can all be squared. The results of that process can all be multiplied by 3, so the domain of is .Let’s start by finding the range of . Since we are squaring , its outputs are all non-negative. However, the maximum absolute value of is 1, so the maximum value of is 1. Since ranges through all values in , ranges through all values in . Multiplying by 3 results in a vertical stretch by a factor of 3, so the range of is .
- (c)
The domain of is all real numbers, and therefore, all outputs of can be plugged into , meaning that the domain of is all real numbers, .Since increases from -1 to 1 as goes from to , increases from to as goes from to . Since going from to takes us through the entire range of , the range of is . Recall that is an odd function, meaning . Therefore, another way to write the range of is .
Finding Zeros of Trigonometric Funtions.
Now we can recognize that we have a difference of squares, so we have the following: Now we can set each factor on the left equal to zero. So we have After simplifying them a bit, we have Because these are famous values of , we can find values for without using inverse trigonometry. The solution to is and the solution to is . Note that there are other -values for which , but these are the only ones for which .
If we wanted to find those other -values, we could add or subtract multiples of to our solutions in the interval . This is because our function is periodic with period . A complete list of all the zeros of would then be and for all integers . The notation with the means that for any integer value of (), gives us another zero of the function , and likewise for .
Thankfully the left-hand side can factor nicely, so we have
Now we set each factor equal to zero: After simplifying them a bit, we have is outside of the range of so we get no solutions to our equation from that factor. Our only concern is the famous value where . This happens to occur in two places for , so our answer is .
A complete list of all the zeros of would be and for all integers .
But we know all the solutions to this equation! If or , then . In fact, we know that all solutions are and for all integers .
However, to solve our original equation of in terms of , we need to undo our substitution: if , then . Dividing our solutions in terms of by 2, we obtain the solutions and for all integers . This is a complete list of zeros of our function .
It now remains to see which of these zeros are in the interval . A surefire way to do this is to plug in different values for . Plugging in for gives us a negative, which isn’t in the interval , so we plug in 0 and 1 for to obtain and . If we plug in 2 for , we obtain , which is larger than , and so is not in the correct interval. All other values of will similarly produce numbers outside . Therefore, the only zeros in of the form are and . Likewise, the only solutions in the interval of the form are and . Therefore, a list of the zeros of in the interval is given by , , , and .
The previous example should seem quite complicated, so let’s try and summarize what we did. First, we isolated the trigonometric function on one side of the equation. Then, we made a substitution to simplify the equation in into an equation in we knew how to solve. Upon solving, we needed to undo the substitution. The twist here is that we needed to undo the substitution for all the solutions to the -equation, then see which of them were in .
But we know all the solutions to this equation! If or , then . In fact, we know that all solutions are and for all integers .
However, to solve our original equation of in terms of , we need to undo our substitution: if , then . Undoing our substitution by subtracting and dividing by , we obtain the solutions and for all integers . This is a complete list of zeros of our function .
It now remains to see which of these zeros are in the interval . Going through all the relevant values of , we find that , , , , , and are the only zeros in the correct interval.
Average Rate of Change with Trigonometry
We can still find average rate of change with trigonometric functions, but because they are periodic there can be some interesting results.
- (a)
- Find .
- (b)
- Find .
- (a)
- First we substitute in our trig values: Next, we simplify our fractions: Therefore,
- (b)
- First we substitute in our trig values: Next, we simplify: We can see that we have in the numerator, which means that our average rate of change will be . This is because we picked -values with the same -value. Even though we had positive and negative rates of change at some point between and , both endpoints of the interval have the same -value, so on average there was no change.