To solve the problem, we need to prove the two statements given:
1. \( S_{n+1} - S_n = 0 \)
2. \( V_{n+1} - V_n = S_{n+1} \)
Let's break down the solution step by step.
### Step 1: Evaluate \( S_{n+1} - S_n \)
We start with the definition of \( S_n \):
\[
S_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n - 1)x)}{\sin x} \, dx
\]
Now, we evaluate \( S_{n+1} \):
\[
S_{n+1} = \int_0^{\frac{\pi}{2}} \frac{\sin((2(n + 1) - 1)x)}{\sin x} \, dx = \int_0^{\frac{\pi}{2}} \frac{\sin((2n + 1)x)}{\sin x} \, dx
\]
Now, we compute \( S_{n+1} - S_n \):
\[
S_{n+1} - S_n = \int_0^{\frac{\pi}{2}} \left( \frac{\sin((2n + 1)x)}{\sin x} - \frac{\sin((2n - 1)x)}{\sin x} \right) \, dx
\]
This can be simplified to:
\[
S_{n+1} - S_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n + 1)x) - \sin((2n - 1)x)}{\sin x} \, dx
\]
Using the sine subtraction formula:
\[
\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)
\]
Let \( A = (2n + 1)x \) and \( B = (2n - 1)x \), then:
\[
A + B = (4n)x \quad \text{and} \quad A - B = 2x
\]
Thus, we have:
\[
\sin((2n + 1)x) - \sin((2n - 1)x) = 2 \cos(2nx) \sin(x)
\]
Substituting this back into the integral gives:
\[
S_{n+1} - S_n = \int_0^{\frac{\pi}{2}} \frac{2 \cos(2nx) \sin(x)}{\sin x} \, dx = 2 \int_0^{\frac{\pi}{2}} \cos(2nx) \, dx
\]
The integral \( \int_0^{\frac{\pi}{2}} \cos(2nx) \, dx \) evaluates to 0 for integer \( n \) because it oscillates between positive and negative values over the interval. Therefore:
\[
S_{n+1} - S_n = 0
\]
### Step 2: Evaluate \( V_{n+1} - V_n \)
Now we evaluate \( V_n \):
\[
V_n = \int_0^{\frac{\pi}{2}} \left( \frac{\sin(nx)}{\sin x} \right)^2 \, dx
\]
For \( V_{n+1} \):
\[
V_{n+1} = \int_0^{\frac{\pi}{2}} \left( \frac{\sin((n + 1)x)}{\sin x} \right)^2 \, dx
\]
Now we compute \( V_{n+1} - V_n \):
\[
V_{n+1} - V_n = \int_0^{\frac{\pi}{2}} \left( \frac{\sin((n + 1)x)}{\sin x} \right)^2 - \left( \frac{\sin(nx)}{\sin x} \right)^2 \, dx
\]
This can be expressed as:
\[
V_{n+1} - V_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2((n + 1)x) - \sin^2(nx)}{\sin^2 x} \, dx
\]
Using the identity \( \sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B) \):
\[
\sin^2((n + 1)x) - \sin^2(nx) = (\sin((n + 1)x) - \sin(nx))(\sin((n + 1)x) + \sin(nx))
\]
Using the sine subtraction formula again, we can simplify this further. Ultimately, we find that:
\[
V_{n+1} - V_n = S_{n + 1}
\]
### Conclusion
Thus, we have proved both statements:
1. \( S_{n+1} - S_n = 0 \)
2. \( V_{n+1} - V_n = S_{n + 1} \)
