In a triangle ABC, the value of sin A + sin B | vedclass

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(B) In $\Delta ABC,$ we have $A + B + C = 180^\circ.$ We need to evaluate $\sin A + \sin B + \sin C.$ Using the sum-to-product formula $\sin A + \sin

Vedclass · Accepted Answer

$4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

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(B) In $\Delta ABC,$ we have $A + B + C = 180^\circ.$ We need to evaluate $\sin A + \sin B + \sin C.$ Using the sum-to-product formula $\sin A + \sin B = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2},$ we get: $\sin A + \sin B + \sin C = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + 2\sin \frac{C}{2} \cos \frac{C}{2}$ Since $A+B = 180^\circ - C,$ then $\frac{A+B}{2} = 90^\circ - \frac{C}{2}.$ Thus,$\sin \frac{A+B}{2} = \cos \frac{C}{2}.$ Substituting this into the expression: $= 2\cos \frac{C}{2} \cos \frac{A-B}{2} + 2\sin \frac{C}{2} \cos \frac{C}{2}$ $= 2\cos \frac{C}{2} \left( \cos \frac{A-B}{2} + \sin \frac{C}{2} \right)$ Since $\sin \frac{C}{2} = \cos \frac{A+B}{2},$ we have: $= 2\cos \frac{C}{2} \left( \cos \frac{A-B}{2} + \cos \frac{A+B}{2} \right)$ Using the formula $\cos(x-y) + \cos(x+y) = 2\cos x \cos y,$ we get: $= 2\cos \frac{C}{2} \left( 2\cos \frac{A}{2} \cos \frac{B}{2} \right) = 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.$ Therefore,the correct option is $B.$

In a triangle $ABC,$ the value of $\sin A + \sin B + \sin C$ is

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