Questions: sum from i=5 to 9 of sin(i * pi/2) =

Transcript text: $\sum_{i=5}^{9} \sin \left(i \frac{\pi}{2}\right)=$

Solution

Solution Steps

Step 1: Evaluate $ \sin(5 \frac{\pi}{2}) $

We start by calculating $ \sin(5 \frac{\pi}{2}) $. Using the periodic property of the sine function, we find: \[ \sin(5 \frac{\pi}{2}) = \sin\left(5 \frac{\pi}{2} - 2\pi \cdot 2\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

Step 2: Evaluate $ \sin(6 \frac{\pi}{2}) $

Next, we evaluate $ \sin(6 \frac{\pi}{2}) $: \[ \sin(6 \frac{\pi}{2}) = \sin(3\pi) = 0 \]

Step 3: Evaluate $ \sin(7 \frac{\pi}{2}) $

Now, we calculate $ \sin(7 \frac{\pi}{2}) $: \[ \sin(7 \frac{\pi}{2}) = \sin\left(7 \frac{\pi}{2} - 2\pi \cdot 3\right) = \sin\left(\frac{\pi}{2}\right) = -1 \]

Step 4: Evaluate $ \sin(8 \frac{\pi}{2}) $

Next, we evaluate $ \sin(8 \frac{\pi}{2}) $: \[ \sin(8 \frac{\pi}{2}) = \sin(4\pi) = 0 \]

Step 5: Evaluate $ \sin(9 \frac{\pi}{2}) $

Finally, we calculate $ \sin(9 \frac{\pi}{2}) $: \[ \sin(9 \frac{\pi}{2}) = \sin\left(9 \frac{\pi}{2} - 2\pi \cdot 4\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

Step 6: Sum the Results

Now, we sum all the evaluated sine values: \[ \sum_{i=5}^{9} \sin\left(i \frac{\pi}{2}\right) = 1 + 0 - 1 + 0 + 1 = 1 \]