The value of the sin 1^{circ} + sin 2^{circ} | vedclass

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Question

(A) The given expression is $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 359^{\circ}$.We know that $\sin(360^{\circ} - 	heta) = -\sin 	heta$

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$0$

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(A) The given expression is $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 359^{\circ}$.We know that $\sin(360^{\circ} - 	heta) = -\sin 	heta$.Thus,$\sin 359^{\circ} = \sin(360^{\circ} - 1^{\circ}) = -\sin 1^{\circ}$.Similarly,$\sin 358^{\circ} = -\sin 2^{\circ}$,and so on.We can pair the terms as $(\sin 1^{\circ} + \sin 359^{\circ}) + (\sin 2^{\circ} + \sin 358^{\circ}) + \ldots + (\sin 179^{\circ} + \sin 181^{\circ}) + \sin 180^{\circ}$.Since $\sin(180^{\circ} + 	heta) = -\sin 	heta$,we have $\sin 181^{\circ} = -\sin 1^{\circ}$,$\sin 182^{\circ} = -\sin 2^{\circ}$,etc.Each pair sums to $0$,and $\sin 180^{\circ} = 0$.Therefore,the total sum is $0$.

The value of the $\sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 359^{\circ}$ is equal to

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