If $A, B, C, D$ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k$ , then show that the value of
$4 sin \frac{A}{2} + 3 sin \frac{B}{2} + 2 sin \frac{C}{2} + sin \frac{D}{2}$ is equal to $2 \sqrt{1 + k}$

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Solution

$A < B < C < D$ are in ascending order.....(1)
$sin A = sin B = sin C = sin D = k$ (+ive)......(2)
$∴ B = π - A, C = 2 π + A, D = 3 π - A$
We have chosen the values so as to satisfy the conditions (1) and (2)
$E = 4 sin \frac{A}{2} + 3 sin \frac{π - A}{2} + 2 sin \frac{2 π + A}{2} + sin \frac{3 π - A}{2} = 4 sin \frac{A}{2} + 3 cos \frac{A}{2} - 2 sin \frac{A}{2} - cos \frac{A}{2}$
$= 2 (sin \frac{A}{2} + cos \frac{A}{2}) = 2 \sqrt{1 + sin A} = \sqrt{1 + k}$

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