If A+B+C=60^{circ},then cos (30^{circ}-A)+cos | vedclass

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(A) Given $A+B+C=60^{\circ}$,we have $\sin(A+B+C) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2} = \cos(30^{\circ})$.Let $S = \cos(30^{\circ}-A)+\cos(30^{\ci

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

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(A) Given $A+B+C=60^{\circ}$,we have $\sin(A+B+C) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2} = \cos(30^{\circ})$.Let $S = \cos(30^{\circ}-A)+\cos(30^{\circ}-B)+\cos(30^{\circ}-C)+\cos(30^{\circ})$.Using the sum-to-product formula $\cos X + \cos Y = 2 \cos \frac{X+Y}{2} \cos \frac{X-Y}{2}$:$S = 2 \cos \left(30^{\circ}-\frac{A+B}{2}\right) \cos \left(\frac{B-A}{2}\right) + 2 \cos \left(30^{\circ}-\frac{C}{2}\right) \cos \left(\frac{C}{2}\right)$.Since $A+B = 60^{\circ}-C$,then $30^{\circ}-\frac{A+B}{2} = 30^{\circ}-\frac{60^{\circ}-C}{2} = \frac{C}{2}$.$S = 2 \cos \frac{C}{2} \cos \frac{B-A}{2} + 2 \cos \left(30^{\circ}-\frac{C}{2}\right) \cos \frac{C}{2}$.$S = 2 \cos \frac{C}{2} \left[ \cos \frac{B-A}{2} + \cos \left(30^{\circ}-\frac{C}{2}\right) \right]$.Using $30^{\circ}-\frac{C}{2} = 30^{\circ}-\frac{60^{\circ}-A-B}{2} = \frac{A+B}{2}$.$S = 2 \cos \frac{C}{2} \left[ \cos \frac{B-A}{2} + \cos \frac{A+B}{2} \right] = 2 \cos \frac{C}{2} \left[ 2 \cos \frac{B}{2} \cos \frac{A}{2} \right]$.$S = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.

If $A+B+C=60^{\circ}$,then $\cos (30^{\circ}-A)+\cos (30^{\circ}-B)+\cos (30^{\circ}-C)+\sin (A+B+C) = $

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