Question

complete
the table
for the
missing
values of
f(θ)
θ | f(θ)
0 | 0
π/4 | □
π/2 | 0
3π/4 | □
π | 0

Explanation:

Assuming \( f(\theta) \) is a sine or cosine - like function, but from the zeros at \( \theta = 0,\frac{\pi}{2},\pi \), a possible function is \( f(\theta)=\sin(2\theta) \) (since the period - related behavior: the zeros of \( \sin(2\theta) \) are at \( 2\theta = n\pi\) or \( \theta=\frac{n\pi}{2},n\in\mathbb{Z} \), which matches the given zeros at \( 0,\frac{\pi}{2},\pi \)).

Step 1: Calculate \( f(\frac{\pi}{4}) \)

If \( f(\theta)=\sin(2\theta) \), substitute \( \theta = \frac{\pi}{4} \) into the function.
We know that \( 2\theta=2\times\frac{\pi}{4}=\frac{\pi}{2} \), and \( \sin(\frac{\pi}{2}) = 1 \). So \( f(\frac{\pi}{4})=\sin(2\times\frac{\pi}{4})=\sin(\frac{\pi}{2}) = 1 \).

Step 2: Calculate \( f(\frac{3\pi}{4}) \)

Substitute \( \theta=\frac{3\pi}{4} \) into \( f(\theta)=\sin(2\theta) \).
We have \( 2\theta = 2\times\frac{3\pi}{4}=\frac{3\pi}{2} \), and \( \sin(\frac{3\pi}{2})=- 1 \). So \( f(\frac{3\pi}{4})=\sin(2\times\frac{3\pi}{4})=\sin(\frac{3\pi}{2})=-1 \).

Answer:

For \( \theta = \frac{\pi}{4} \), \( f(\theta)=1 \); for \( \theta=\frac{3\pi}{4} \), \( f(\theta)=- 1 \)

(If we assume a different function, for example, \( f(\theta)=\cos(2\theta - \frac{\pi}{2})=\sin(2\theta) \) we still get the same result. Another possible function could be a triangular wave, but the sinusoidal function is a common choice here. If we consider \( f(\theta)=\sin(2\theta) \), the values are as calculated above.)