If `sum_(i=1)^(n) cos theta_(i)=n`, then the value of `sum_(i=1)^(n) sin theta_(i)`.

Answer

To solve the problem, we start with the given condition:

\[ \sum_{i=1}^{n} \cos \theta_i = n \]

### Step 1: Understand the implication of the summation
The expression \(\sum_{i=1}^{n} \cos \theta_i = n\) means that the sum of the cosine values for \(n\) angles \(\theta_i\) equals \(n\).

### Step 2: Relate the cosine values to their maximum
The maximum value of \(\cos \theta\) is 1. For the sum of \(n\) cosines to equal \(n\), each individual cosine must equal 1. This is because if any \(\cos \theta_i\) were less than 1, the total would be less than \(n\).

Thus, we conclude:

\[ \cos \theta_i = 1 \quad \text{for all } i = 1, 2, \ldots, n \]

### Step 3: Determine the corresponding sine values
The cosine function and sine function are related through the Pythagorean identity:

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Since we have established that \(\cos \theta_i = 1\), we can substitute this into the identity:

\[ \sin^2 \theta_i + 1^2 = 1 \]

This simplifies to:

\[ \sin^2 \theta_i + 1 = 1 \]

### Step 4: Solve for sine
From the equation above, we can isolate \(\sin^2 \theta_i\):

\[ \sin^2 \theta_i = 0 \]

Taking the square root of both sides gives:

\[ \sin \theta_i = 0 \quad \text{for all } i = 1, 2, \ldots, n \]

### Step 5: Calculate the summation of sine values
Now, we can find the summation of the sine values:

\[ \sum_{i=1}^{n} \sin \theta_i = \sum_{i=1}^{n} 0 = 0 \]

### Final Answer
Thus, the value of \(\sum_{i=1}^{n} \sin \theta_i\) is:

\[ \boxed{0} \]

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