If A,b,c,d are the smallest positive angles in ascending order of magnitude their sines equal to the positive quantity K , then $4 sin \frac{A}{2} + 3 sin \frac{B}{2} + 2 sin \frac{C}{2} + sin \frac{D}{2} = 2 \sqrt{1 + k}$ .

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Solution

$sin A = sin B = sin C = sin D = K$

$B = π - A = \frac{B}{2} = \frac{π}{2} - \frac{A}{2}$

$C = 2 π + A = \frac{C}{2} = π + \frac{A}{2}$

$D = 3 π - A = \frac{D}{2} = \frac{3 π}{2} - \frac{A}{2}$

Now $4 sin \frac{A}{2} + 3 sin \frac{B}{2} + 2 sin \frac{C}{2} + sin \frac{D}{2}$

$4 sin \frac{A}{2} + 3 cos \frac{A}{2} + 2 sin \frac{A}{2} - cos \frac{A}{2}$

$2 sin \frac{A}{2} + 2 cos \frac{A}{2}$

$2 (sin \frac{A}{2} + cos \frac{A}{2})$

$2 \sqrt{{(sin \frac{A}{2} + cos \frac{A}{2})}^{2}}$

$2 \sqrt{{sin}^{2} \frac{A}{2} + {cos}^{2} \frac{A}{2} + sin A}$

$2 \sqrt{1 + sin A}$

$2 \sqrt{1 + K}$

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