Problem 69 - Precalculus: A Concise Course

Question

Because \(f(t)=\sin t\) is an odd function and \(g(t)=\cos t\) is an even function, what can be said about the function \(h(t)=f(t) g(t) ?\)

Step-by-Step Solution

Answer

The function \(h(t)=f(t)g(t)\) is an odd function.

Step 1: Recall the properties of even and odd functions

An even function follows the rule \(f(-x) = f(x)\). An odd function follows the rule \(f(-x) = -f(x)\).

Step 2: Determine the nature of the newly formed function

Calculate \(h(-t)\), which is equal to \(f(-t)g(-t)\). Since \(f(t)= \sin t\) is an odd function, substitute \(f(-t)\) with \(-f(t)\). And since \(g(t) = \cos t\) is an even function, substitute \(g(-t)\) with \(g(t)\). This gives us \(h(-t) = -f(t)g(t)\), which is equal to \(-h(t)\). Thus, we can conclude that \(h(t)\) is an odd function.

Step 3: Conclusion

The product of an odd function and an even function is an odd function. Therefore, \(h(t)=f(t)g(t)\) is an odd function.