If α , β ∈ ( − 2 π , 0 ) such that ( sin α + sin β ) + s i n β s i n α = 0 and ( sin α + sin β ) s i n β s i n α = − 1 , then λ = lim n → ∞ ( 1 + 2 s i n 2 n β 1 + 2 s i n 2 n α ) .

Question

If α , β ∈ ( − 2 π ​ , 0 ) such that ( sin α + sin β ) + s i n β s i n α ​ = 0 and ( sin α + sin β ) s i n β s i n α ​ = − 1 , then λ = lim n → ∞ ​ ( 1 + 2 s i n 2 n β 1 + 2 s i n 2 n α ​ ) .

TestHub · Accepted Answer

( sin α + sin β ) 2 = 1 ​ ⇒ sin α + sin β = − 1 ​ ​ ​ ⇒ sin α = 2 − 1 ​ sin β = 2 − 1 ​ α = β = − 3 0 ∘

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