<p>The value of \(E = \sin\dfrac{\pi}{n} + \sin\dfrac{3\pi}{n} + \sin\dfrac{5\pi}{n} + \cdots\) to \(n\) terms is:</p>
Step-by-Step Solution
Key Concept: The sum consists of sines of odd multiples of π/n, which exhibit symmetry properties. When n is even, terms pair up to cancel; when n is odd, the series has specific symmetry that results in zero.
<p><strong>Step 1:</strong> Write the series explicitly. We have E = sin(π/n) + sin(3π/n) + sin(5π/n) + ... with n terms total.</p><p><strong>Step 2:</strong> Identify the general term. The k-th term is sin((2k-1)π/n) where k = 1, 2, 3, ..., n.</p><p><strong>Step 3:</strong> Use the symmetry property. Notice that sin((2k-1)π/n) and sin((2(n-k)+1)π/n) = sin((2n-2k+1)π/n) form complementary pairs with respect to π/2.</p><p><strong>Step 4:</strong> Apply the identity sin(π - θ) = sin(θ). For the (n-k+1)-th term: sin((2(n-k+1)-1)π/n) = sin((2n-2k+1)π/n) = sin(π - (2k-1)π/n) = sin((2k-1)π/n).</p><p><strong>Step 5:</strong> However, observe that terms equidistant from the center have the property: the m-th term from the start and m-th term from the end sum using the fact that sin(x) + sin(π-x) = 2sin(x)cos(π-x). But more directly, when we pair terms symmetrically about π/2, they exhibit cancellation.</p><p><strong>Step 6:</strong> The complete analysis shows that consecutive pairs of terms (when arranged by symmetry) cancel out. Specifically, sin((2k-1)π/n) + sin((2(n-k)+1)π/n) evaluates through the sum-to-product formula, yielding zero for the complete pairing.</p><p><strong>Step 7:</strong> Alternatively, use the formula for sum of sines in AP: Σsin(a + (k-1)d) = [sin(nd/2)/sin(d/2)]·sin(a + (n-1)d/2), where a = π/n and d = 2π/n. Here the numerator equals sin(π) = 0.</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B