Question

\( \frac{\sin 1^\circ + \sin 2^\circ + \dots + \sin 89^\circ}{2(\cos 1^\circ + \cos 2^\circ + \dots + \cos 44^\circ) + 1} = \)

• A. 2
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \frac{1}{2} \)
• D. \( \sqrt{2} \)

Hint:

The identity \( 1 + 2\cos \theta + 2\cos 2\theta + \dots + 2\cos n\theta = \frac{\sin(n+1/2)\theta}{\sin(\theta/2)} \) is extremely useful for such series.

Answer 1

The Correct Option is B

Step 1: Simplify Numerator and Denominator:

Let the numerator be \( S \). \[ S = \sum_{k=1}^{89} \sin k^\circ \] Using the property \( \sin(90^\circ - \theta) = \cos\theta \), we can pair terms or rewrite the sum. Alternatively, use the sum of sines formula: \[ \sum_{k=1}^n \sin(kx) = \frac{\sin(nx/2)}{\sin(x/2)} \sin((n+1)x/2) \] For \( n=89, x=1^\circ \): \[ S = \frac{\sin(44.5^\circ)}{\sin(0.5^\circ)} \sin(45^\circ) = \frac{\sin(44.5^\circ)}{\sin(0.5^\circ)} \frac{1}{\sqrt{2}} \] Now let the denominator be \( D \). \[ D = 1 + 2\sum_{k=1}^{44} \cos k^\circ \] Recall the Dirichlet kernel identity or sum of cosines: \[ 1 + 2\sum_{k=1}^n \cos(kx) = \frac{\sin((n+1/2)x)}{\sin(x/2)} \] For \( n=44, x=1^\circ \): \[ D = \frac{\sin((44 + 0.5)^\circ)}{\sin(0.5^\circ)} = \frac{\sin(44.5^\circ)}{\sin(0.5^\circ)} \]
Step 2: Calculate Ratio:

\[ \frac{S}{D} = \frac{ \frac{\sin(44.5^\circ)}{\sin(0.5^\circ)} \frac{1}{\sqrt{2}} }{ \frac{\sin(44.5^\circ)}{\sin(0.5^\circ)} } \] Canceling the common term: \[ \frac{S}{D} = \frac{1}{\sqrt{2}} \]
Step 4: Final Answer:

The value is \( \frac{1}{\sqrt{2}} \).