Suppose that $-90^{\circ}<\theta<90^{\circ}$. Find the sign of the function value $\cos \left(\theta+180^{\circ}\right)$. Choose the correct answer below. Negative Positive

Step 1 ：Suppose that \(-90^{\circ}<\theta<90^{\circ}\). We are asked to find the sign of the function value \(\cos \left(\theta+180^{\circ}\right)\).

Step 2 ：The cosine function has a period of \(360^{\circ}\). This means that \(\cos(\theta + 180^{\circ})\) is the same as \(\cos(\theta - 180^{\circ})\).

Step 3 ：The cosine function is positive in the first and fourth quadrants, and negative in the second and third quadrants.

Step 4 ：If \(-90^{\circ}<\theta<90^{\circ}\), then \(90^{\circ}<\theta+180^{\circ}<270^{\circ}\), which falls in the second and third quadrants. Therefore, we might initially think that \(\cos(\theta + 180^{\circ})\) should be negative.

Step 5 ：However, not all the values of \(\cos(\theta + 180^{\circ})\) are negative for \(-90^{\circ}<\theta<90^{\circ}\). This is because the cosine function is not negative for all values in the second and third quadrants. It is positive near the x-axis in the second quadrant and near the x-axis in the third quadrant.

Step 6 ：Final Answer: The sign of the function value \(\cos \left(\theta+180^{\circ}\right)\) for \(-90^{\circ}<\theta<90^{\circ}\) is not always negative. It can be both negative and positive. Therefore, the correct answer is neither 'Negative' nor 'Positive'. \(\boxed{\text{Neither Negative nor Positive}}\)