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(b) [First analyze the statement to be proved.] BE² = AC × DC - EC² ⇔ BE² + EC² = AC × DC Here the left-hand side is actually BC² by Pythagoras theorem. Therefore, we are...

4sin(x)sin(3x)sin(5x)     = 2(cos(3x-x)-cos(3x+x))sin(5x)     = 2cos(2x)sin(5x) - 2cos(4x)sin(5x)     = (sin(5x+2x)+sin(5x-2x)) - (sin(5x+4x)+sin(5x-4x))     = sin(7x) + sin(3x) - sin(9x) - sin(x)     = -sin(x) + sin(3x) + sin(7x) - sin(9x) sin(75°)sin(45°)     = (cos(75°-45°) - cos(75°+45°))/2     = (cos(30°) - cos(120°))/2     = (√3/2 + 1/2)/2 = (√3...

其實就是計算  ∫_[0 to 2] 3x^2 dx  而已. 不過, 題目說要用 "極限" 方法, 也就是要用 黎曼和. 取點 2n/, 4/n, ..., (2n)/n. 則     Σ_{k = 1 to n} 3[2(k-1)/n]^2．(2/n)       ≦  ∫_[0 to 2] 3x^2 dx = A       ≦ Σ_{k = 1 to n} 3(2k/n)^2．(2/n) 即      Σ_{i = 0 to n-1} 12i^2/n^2．(2...

x = sec(θ) ∴ |x| ≧ 1 ∴ x^2 - 1 = tan^2(θ)    (x^2-1)/x^2 = sin^2(θ) (x^2-2)^2/x^4 = [(x^2-1)/x^2-1/x^2]^2    = (sin^2(θ) -1/x^2)^2    ≦ (max{sin^2(θ),1/x^2})^2 (*)    ≦ 1 關於 (*): a, b 均非負, 則     |a-b| = max{a-b,b-a} ≦ max{a,b} 故     (a-b)^2 ≦ (max{a,b})^2 等號僅成立於 a, b 兩者其一為 0. 若 max{sin^2(θ),1/x^2} = sin^2(θ...

1 + 3sinθcosθ + 2sin²θ - 2sin⁴θ = 0 1 + 3sinθcosθ + 2sin²θ(1 - sin²θ) = 0 1 + 3sinθcosθ + 2sin²θcos²θ = 0 (1 + sinθcosθ)(1 + 2sinθcosθ) = 0 sinθcosθ = -1   or   sinθcosθ = -1/2 2sinθcosθ = -2   or   2sinθcosθ = -1 sin(2θ) = -2 (rejected...

(a) LHS = sin⁴θ + cos⁴θ = sin⁴θ + 2sin²θcos²θ + cos⁴θ - 2sin²θcos²θ = (sin²θ + cos²θ)² - (1/2) 2²sin²θcos²θ = 1² - (1/2) (2sinθcosθ)² = 1 - (1/2...

△ABC ～ △ADE AC/AB = AE/AD 1o/8 = (8+14)/(1o+CD) ∴ CD = 22/(1o/8) - 1o = 7.6 (cm) DE = √(AE^2-AD^2) = √(22^2-17.6^2) = 13.2 EC = √(13.2^2+7.6^2) = √232 圓BCDE半徑 = √232/2 = √58(cm)

in加了底線當然要改，floor 可解作樓層或地板，通常你坐在地上叫做sit on the floor ，所以應該改成on which floor