The value of sin 10^circ + sin 20^circ + sin | vedclass

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(B) Let $S = \sum_{k=1}^{36} \sin(k 	imes 10^\circ)$.We know that $\sin(180^\circ + 	heta) = -\sin 	heta$.Thus,$\sin(180^\circ + 10^\circ) = \sin 1

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$0$

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(B) Let $S = \sum_{k=1}^{36} \sin(k 	imes 10^\circ)$.We know that $\sin(180^\circ + 	heta) = -\sin 	heta$.Thus,$\sin(180^\circ + 10^\circ) = \sin 190^\circ = -\sin 10^\circ$.Similarly,$\sin(180^\circ + 20^\circ) = \sin 200^\circ = -\sin 20^\circ$,and so on up to $\sin(180^\circ + 170^\circ) = \sin 350^\circ = -\sin 170^\circ$.Also,$\sin 180^\circ = 0$ and $\sin 360^\circ = 0$.The sum can be written as $(\sin 10^\circ + \sin 190^\circ) + (\sin 20^\circ + \sin 200^\circ) + \dots + (\sin 170^\circ + \sin 350^\circ) + \sin 180^\circ + \sin 360^\circ$.Since each pair sums to $0$ and $\sin 180^\circ = \sin 360^\circ = 0$,the total sum is $0$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

The value of $\sin 10^\circ + \sin 20^\circ + \sin 30^\circ + \dots + \sin 360^\circ$ is

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