回覆 3# zj0209 的帖子
第 12 題
sinθ = a > 0
cosθ = b > 0
sinψ = c > 0
cosψ = d > 0
a^2 + b^2 = c^2 + d^2 = 1
原題改為 a^2024/d^2022 + b^2024/c^2022 = 1,求 a^2023 - d^2023
(a^2024/d^2022 + b^2024/c^2022)(d^2022/a^2020 + c^2022/b^2020) ≧ (a^2 + b^2)^2 = 1
等號成立於 (a/d)^2024) = (b/c)^2024,ac = bd,θ + ψ = π/2
此時 d^2022/a^2020 + c^2022/b^2020 = 1
所求 = (sinθ)^2023 - (cosψ)^2023 = 0