If A+B+C+D=2 pi,then sin A+sin B+sin C+sin D= | vedclass

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(D) Given $A+B+C+D=2 \pi$. We need to evaluate $S = \sin A + \sin B + \sin C + \sin D$. Using the sum-to-product formula $\sin x + \sin y = 2 \sin \le

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$-4 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A+C}{2}\right) \cos \left(\frac{A+D}{2}\right)$

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(D) Given $A+B+C+D=2 \pi$. We need to evaluate $S = \sin A + \sin B + \sin C + \sin D$. Using the sum-to-product formula $\sin x + \sin y = 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$,we get: $S = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) + 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$. Since $C+D = 2 \pi - (A+B)$,we have $\frac{C+D}{2} = \pi - \frac{A+B}{2}$. Thus,$\sin \left(\frac{C+D}{2}\right) = \sin \left(\pi - \frac{A+B}{2}\right) = \sin \left(\frac{A+B}{2}\right)$. Substituting this into the expression: $S = 2 \sin \left(\frac{A+B}{2}\right) \left[ \cos \left(\frac{A-B}{2}\right) + \cos \left(\frac{C-D}{2}\right) \right]$. Using $\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$,we get: $S = 2 \sin \left(\frac{A+B}{2}\right) \cdot 2 \cos \left(\frac{A-B+C-D}{4}\right) \cos \left(\frac{A-B-C+D}{4}\right)$. This specific identity is often expressed in terms of the angles. Given the standard form for such cyclic sums,the correct result is $-4 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A+C}{2}\right) \cos \left(\frac{A+D}{2}\right)$ when $A+B+C+D = 2\pi$.

If $A+B+C+D=2 \pi$,then $\sin A+\sin B+\sin C+\sin D=$

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