frac{sin 1^{circ}+sin 2^{circ}+ldots+sin 89^{ | vedclass

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Question

(B) Let $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 89^{\circ}$.Using the sum formula for sines in arithmetic progression,$\sum_{k=1}^{n} \si

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(B) Let $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 89^{\circ}$.Using the sum formula for sines in arithmetic progression,$\sum_{k=1}^{n} \sin(k	heta) = \frac{\sin(n	heta/2) \sin((n+1)	heta/2)}{\sin(	heta/2)}$.Here $n = 89$ and $	heta = 1^{\circ}$,so $S = \frac{\sin(89^{\circ}/2) \sin(90^{\circ}/2)}{\sin(0.5^{\circ})} = \frac{\sin(44.5^{\circ}) \sin(45^{\circ})}{\sin(0.5^{\circ})}$.Now consider the denominator $D = 2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1$.Using the sum formula $\sum_{k=1}^{n} \cos(k	heta) = \frac{\sin(n	heta/2) \cos((n+1)	heta/2)}{\sin(	heta/2)}$,we have $\sum_{k=1}^{44} \cos(k^{\circ}) = \frac{\sin(44^{\circ}/2) \cos(45^{\circ}/2)}{\sin(0.5^{\circ})} = \frac{\sin(22^{\circ}) \cos(22.5^{\circ})}{\sin(0.5^{\circ})}$.Alternatively,note that $\sin 1^{\circ} + \ldots + \sin 89^{\circ} = (\sin 1^{\circ} + \sin 89^{\circ}) + \ldots + \sin 45^{\circ} = 2 \sin 45^{\circ} \cos 44^{\circ} + 2 \sin 45^{\circ} \cos 43^{\circ} + \ldots + \sin 45^{\circ}$.This simplifies to $\sqrt{2}(\cos 44^{\circ} + \cos 43^{\circ} + \ldots + \cos 1^{\circ}) + \frac{1}{\sqrt{2}}$.Multiplying by $\sqrt{2}$,we get $\sqrt{2} 	imes S = 2(\cos 1^{\circ} + \ldots + \cos 44^{\circ}) + 1$.Thus,$\frac{S}{2(\cos 1^{\circ} + \ldots + \cos 44^{\circ}) + 1} = \frac{1}{\sqrt{2}}$.

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

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