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Period of Trigonometric Ratios

If α·β, Y , δ...

Question

If $α . β, γ, δ$ are the smallest positive angles in ascending order of magnitude which have their sines equal to positive quantity k, then the value of $4 s i n (\frac{α}{2}) + 3 s i n (\frac{β}{2}) + 2 s i n (\frac{γ}{2}) + s i n (\frac{δ}{2})$ is not equal to

$2 \sqrt{1 - k}$

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$2 \sqrt{1 + k}$

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$2 \sqrt{k}$

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Solution

The correct options are
A
$2 \sqrt{1 - k}$

C
$2 \sqrt{k}$

D
2k

Given $α < β < γ < δ$ also $s i n α = s i n β = s i n γ = s i n δ = k$ and $α, β, γ, δ$ are the smallest positive angles.
$∴ β = π - α . γ = 2 π + α, δ = 3 π - α$
Substituting the values in given expression
We get $2 (s i n \frac{α}{2} + c o s \frac{α}{2}) = 2 \sqrt{1 + s i n α} = 2 \sqrt{1 + k}$

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