Problem 147 - Algebra and Trigonometry Real Mathematics, Real People

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

Step 1: Understanding the properties of odd and even functions

An even function \(f(x)\) satisfies \(f(x) = f(-x)\) for all \(x\) in the function’s domain. An odd function \(f(x)\) satisfies \(f(-x) = -f(x)\) for all \(x\) in the function’s domain. The function \(h(t)\) is expressed as a product of an odd function \(f(t) = \sin(t)\) and an even function \(g(t) = \cos(t)\).

Step 2: Verifying the property of function \(h(t)\)

Replace \(t\) with \(-t\) in \(h(t)\). If \(h(t)\) is even, then \(h(t) = h(-t)\), if \(h(t)\) is odd, then \(h(-t)=-h(t)\), if neither condition is satisfied, the function is neither even nor odd. So, \(h(-t)=f(-t)g(-t)=(-sin(t))cos(t)=-f(t)g(t)=-h(t)\). Hence \(h(t)\) is an odd function.

Step 3: Conclusion

From the result obtained in step 2, the product of an odd function and an even function, in this case \(h(t) = f(t)g(t) = sin(t)cos(t)\), is an odd function.