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

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Question

(B) Let $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 89^{\circ}$.We can pair terms as $(\sin 1^{\circ} + \sin 89^{\circ}) + (\sin 2^{\circ} +

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 89^{\circ}$.We can pair terms as $(\sin 1^{\circ} + \sin 89^{\circ}) + (\sin 2^{\circ} + \sin 88^{\circ}) + \ldots + (\sin 44^{\circ} + \sin 46^{\circ}) + \sin 45^{\circ}$.Using the identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we have $\sin k^{\circ} + \sin(90-k)^{\circ} = 2 \sin 45^{\circ} \cos(\frac{90-2k}{2}) = 2 \cdot \frac{1}{\sqrt{2}} \cos(45-k)^{\circ} = \sqrt{2} \cos(45-k)^{\circ}$.Alternatively,$\sin k^{\circ} + \sin(90-k)^{\circ} = \sin k^{\circ} + \cos k^{\circ}$.Actually,$\sin k^{\circ} + \sin(90-k)^{\circ} = 2 \sin 45^{\circ} \cos(k-45)^{\circ} = \sqrt{2} \cos(k-45)^{\circ}$.Using $\sin A + \sin B = 2 \sin 45^{\circ} \cos(\frac{A-B}{2})$,the numerator becomes $\sqrt{2} \cos 44^{\circ} + \sqrt{2} \cos 43^{\circ} + \ldots + \sqrt{2} \cos 1^{\circ} + \sin 45^{\circ}$.Numerator $= \sqrt{2} (\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + \frac{1}{\sqrt{2}}$.Numerator $= \frac{1}{\sqrt{2}} [2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1]$.Thus,the expression simplifies to $\frac{1}{\sqrt{2}}$.

$\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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