The value of the expression frac{2(sin 1^{cir | vedclass

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Question

(A) Let $S = \sin 1^{\circ} + \sin 2^{\circ} + \dots + \sin 89^{\circ}$.Using the identity $\sin 	heta + \sin(90^{\circ} - 	heta) = 2 \sin 45^{\circ

Vedclass · Accepted Answer

$\sqrt{2}$

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(A) Let $S = \sin 1^{\circ} + \sin 2^{\circ} + \dots + \sin 89^{\circ}$.Using the identity $\sin 	heta + \sin(90^{\circ} - 	heta) = 2 \sin 45^{\circ} \cos(	heta - 45^{\circ}) = \sqrt{2} \cos(	heta - 45^{\circ})$,we can pair terms: $(\sin 1^{\circ} + \sin 89^{\circ}) + (\sin 2^{\circ} + \sin 88^{\circ}) + \dots + (\sin 44^{\circ} + \sin 46^{\circ}) + \sin 45^{\circ}$.Each pair $(\sin 	heta + \sin(90^{\circ} - 	heta)) = 2 \sin 45^{\circ} \cos(45^{\circ} - 	heta) = \sqrt{2} \cos(45^{\circ} - 	heta)$.Thus,the numerator is $2 	imes [\sqrt{2}(\cos 44^{\circ} + \cos 43^{\circ} + \dots + \cos 1^{\circ}) + \sin 45^{\circ}]$.Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,the numerator becomes $2\sqrt{2}(\cos 44^{\circ} + \dots + \cos 1^{\circ}) + 2(\frac{1}{\sqrt{2}}) = 2\sqrt{2}(\cos 44^{\circ} + \dots + \cos 1^{\circ}) + \sqrt{2}$.Factoring out $\sqrt{2}$,we get $\sqrt{2} [2(\cos 1^{\circ} + \dots + \cos 44^{\circ}) + 1]$.Dividing by the denominator $2(\cos 1^{\circ} + \dots + \cos 44^{\circ}) + 1$,the result is $\sqrt{2}$.

The value of the expression $\frac{2(\sin 1^{\circ} + \sin 2^{\circ} + \sin 3^{\circ} + \dots + \sin 89^{\circ})}{2(\cos 1^{\circ} + \cos 2^{\circ} + \dots + \cos 44^{\circ}) + 1}$ equals

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