If A+B+C=frac{pi}{2},then sqrt{2} cos left(fr | vedclass

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(A) Given $A+B+C=\frac{\pi}{2}$. We need to evaluate $S = \sqrt{2} \cos \left(\frac{\pi}{4}-A\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-B\right)+\sqrt{

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $A+B+C=\frac{\pi}{2}$. We need to evaluate $S = \sqrt{2} \cos \left(\frac{\pi}{4}-A\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-B\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-C\right)+1$. Note that $1 = \sqrt{2} \cos \left(\frac{\pi}{4}\right)$. So,$S = \sqrt{2} [\cos(\frac{\pi}{4}-A) + \cos(\frac{\pi}{4}-B) + \cos(\frac{\pi}{4}-C) + \cos(\frac{\pi}{4})]$. Using $\cos X + \cos Y = 2 \cos \frac{X+Y}{2} \cos \frac{X-Y}{2}$: $S = \sqrt{2} [2 \cos(\frac{\pi/2 - (A+B)}{2}) \cos(\frac{B-A}{2}) + 2 \cos(\frac{\pi/2 - C}{2}) \cos(\frac{C}{2})]$. Since $A+B = \frac{\pi}{2}-C$,then $\frac{\pi/2 - (A+B)}{2} = \frac{C}{2}$. $S = 2\sqrt{2} \cos(\frac{C}{2}) [\cos(\frac{B-A}{2}) + \cos(\frac{\pi/4 - C/2}{2})]$. Using $\frac{\pi}{4} - \frac{C}{2} = \frac{A+B}{2}$,we simplify the expression to $4\sqrt{2} \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.

If $A+B+C=\frac{\pi}{2}$,then $\sqrt{2} \cos \left(\frac{\pi}{4}-A\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-B\right)+\sqrt{2} \cos \left(\frac{\pi}{4}-C\right)+1=$

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