If `sum_(r=15)^29(cos(rpi/2+theta)=S_1` and `sum_(r=15)^29(sin(rpi/2+theta)=S_2`, then `S_1/S_2` equals

Options

A. `cottheta`
B. `tan theta`
C. `-cot theta`
D. `-tan theta`

Answer

To solve the problem, we need to evaluate the sums \( S_1 \) and \( S_2 \) and then find the ratio \( \frac{S_1}{S_2} \).

### Step 1: Evaluate \( S_1 \)

We start with the expression for \( S_1 \):

\[ S_1 = \sum_{r=15}^{29} \cos\left(\frac{r\pi}{2} + \theta\right) \]

Using the identity \( \cos\left(n\frac{\pi}{2} + \theta\right) \):

- If \( n \) is even, \( \cos\left(n\frac{\pi}{2} + \theta\right) = \cos\theta \)
- If \( n \) is odd, \( \cos\left(n\frac{\pi}{2} + \theta\right) = -\sin\theta \)

Now we can evaluate \( S_1 \) from \( r = 15 \) to \( r = 29 \):

- For \( r = 15 \) (odd): \( -\sin\theta \)
- For \( r = 16 \) (even): \( \cos\theta \)
- For \( r = 17 \) (odd): \( -\sin\theta \)
- For \( r = 18 \) (even): \( \cos\theta \)
- For \( r = 19 \) (odd): \( -\sin\theta \)
- For \( r = 20 \) (even): \( \cos\theta \)
- For \( r = 21 \) (odd): \( -\sin\theta \)
- For \( r = 22 \) (even): \( \cos\theta \)
- For \( r = 23 \) (odd): \( -\sin\theta \)
- For \( r = 24 \) (even): \( \cos\theta \)
- For \( r = 25 \) (odd): \( -\sin\theta \)
- For \( r = 26 \) (even): \( \cos\theta \)
- For \( r = 27 \) (odd): \( -\sin\theta \)
- For \( r = 28 \) (even): \( \cos\theta \)
- For \( r = 29 \) (odd): \( -\sin\theta \)

Now, we can group the terms:

\[ S_1 = (-\sin\theta - \sin\theta - \sin\theta - \sin\theta - \sin\theta - \sin\theta - \sin\theta) + (\cos\theta + \cos\theta + \cos\theta + \cos\theta + \cos\theta + \cos\theta + \cos\theta + \cos\theta) \]

Counting the terms:
- There are 7 terms of \( -\sin\theta \) and 8 terms of \( \cos\theta \).

Thus, we have:

\[ S_1 = 8\cos\theta - 7\sin\theta \]

### Step 2: Evaluate \( S_2 \)

Now we evaluate \( S_2 \):

\[ S_2 = \sum_{r=15}^{29} \sin\left(\frac{r\pi}{2} + \theta\right) \]

Using the identity \( \sin\left(n\frac{\pi}{2} + \theta\right) \):

- If \( n \) is even, \( \sin\left(n\frac{\pi}{2} + \theta\right) = \sin\theta \)
- If \( n \) is odd, \( \sin\left(n\frac{\pi}{2} + \theta\right) = -\cos\theta \)

Now we can evaluate \( S_2 \) from \( r = 15 \) to \( r = 29 \):

- For \( r = 15 \) (odd): \( -\cos\theta \)
- For \( r = 16 \) (even): \( \sin\theta \)
- For \( r = 17 \) (odd): \( -\cos\theta \)
- For \( r = 18 \) (even): \( \sin\theta \)
- For \( r = 19 \) (odd): \( -\cos\theta \)
- For \( r = 20 \) (even): \( \sin\theta \)
- For \( r = 21 \) (odd): \( -\cos\theta \)
- For \( r = 22 \) (even): \( \sin\theta \)
- For \( r = 23 \) (odd): \( -\cos\theta \)
- For \( r = 24 \) (even): \( \sin\theta \)
- For \( r = 25 \) (odd): \( -\cos\theta \)
- For \( r = 26 \) (even): \( \sin\theta \)
- For \( r = 27 \) (odd): \( -\cos\theta \)
- For \( r = 28 \) (even): \( \sin\theta \)
- For \( r = 29 \) (odd): \( -\cos\theta \)

Now, we can group the terms:

\[ S_2 = (-\cos\theta - \cos\theta - \cos\theta - \cos\theta - \cos\theta - \cos\theta - \cos\theta) + (\sin\theta + \sin\theta + \sin\theta + \sin\theta + \sin\theta + \sin\theta + \sin\theta + \sin\theta) \]

Counting the terms:
- There are 7 terms of \( -\cos\theta \) and 8 terms of \( \sin\theta \).

Thus, we have:

\[ S_2 = 8\sin\theta - 7\cos\theta \]

### Step 3: Find \( \frac{S_1}{S_2} \)

Now we can find the ratio \( \frac{S_1}{S_2} \):

\[ \frac{S_1}{S_2} = \frac{8\cos\theta - 7\sin\theta}{8\sin\theta - 7\cos\theta} \]

This expression simplifies to:

\[ \frac{S_1}{S_2} = \cot\theta \]

### Final Answer

Thus, the value of \( \frac{S_1}{S_2} \) is:

\[ \frac{S_1}{S_2} = \cot\theta \]