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evaluate-P-n-a-sin-a-sin-2-a-sin-2-2-a-sin-2-n-with-a-1-




Question Number 191229 by mr W last updated on 22/Apr/23
evaluate  P_n =(a+sin θ)(a+sin (θ/2))(a+sin (θ/2^2 ))...(a+sin (θ/2^n ))  with ∣a∣≤1
evaluatePn=(a+sinθ)(a+sinθ2)(a+sinθ22)(a+sinθ2n)witha∣⩽1

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