If α, β, γ, δ are smallest positive angle in ascending order of magnitude which have their sines equal to positive quantity λ , then if 4 sin ⁡ α 2 + 3 sin ⁡ β 2 + 2 sin ⁡ γ 2 + sin ⁡ δ 2 = 2 1 + λ then λ = ?Options-(1) sin ⁡ α(2) sin ⁡ β(3) sin ⁡ γ(4) sin ⁡ δ

Question

If   α,   β,   γ,   δ are smallest positive angle in ascending order of magnitude which have their sines equal to positive quantity   λ , then if  4  sin  ⁡      α    2    +  3  sin  ⁡      β    2    +  2  sin  ⁡      γ    2    +  sin  ⁡      δ    2    =  2      1    +    λ  then   λ = ?Options-(1)  sin  ⁡  α(2)  sin  ⁡  β(3)  sin  ⁡  γ(4)  sin  ⁡  δ

percolarse2rzd · Accepted Answer

See we have ascending order for positive sines as   a  ,  π  −  a  ,  2  π  +  a  ,  3  π  −  a as sine is positive only in first two quadrants .Now these are values of   α  ,  β  ,  γ  ,  δ plugging in and simplifying we get   sin  ⁡  (  α      /    2  )  +  cos  ⁡  (  α      /    2  )  =      1    +    λ   thus squaring we get   λ  =  sin  ⁡  (  α  )

If α , β , γ , δ<!-- δ

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