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三角函數的問題

發問:

16 (a) Prove that tan(3π/8)=1+√2. (b) Using (a),prove that tan(π/24)=√6+√2-√3-2.

最佳解答:

a) Consider tan (3π/4) = -1 tan [2(3π/8)] = -1 2tan (3π/8)/[1 - tan2 (3π/8)] = -1 1 - tan2 (3π/8) = -2tan (3π/8) tan2 (3π/8) - 2tan (3π/8) - 1 = 0 tan (3π/8) = (2 +/- √8)/2 = 1 +/- √2 Since tan (3π/8) > 1: tan (3π/8) = 1 + √2 b) Using the fact that 3π/8 - π/24 = π/3: tan (3π/8 - π/24) = tan π/3 = √3 [tan (3π/8) - tan (π/24)]/[1 + tan (3π/8) tan (π/24)] = √3 [1 + √2 - tan (π/24)]/[1 + (1 + √2) tan (π/24)] = √3 1 + √2 - tan (π/24) = √3 [1 + (1 + √2) tan (π/24)] 1 + √2 - tan (π/24) = √3 + √3(1 + √2) tan (π/24) (√3 + √6 + 1) tan (π/24) = 1 + √2 - √3 tan (π/24) = (1 + √2 - √3)/(√3 + √6 + 1) = [(1 + √2 - √3)(√3 + √6 - 1)]/[(√3 + √6 + 1)(√3 + √6 - 1)] = (4√3 + 2√6 - 4√2 - 4)/(8 + 6√2) = (2√3 + √6 - 2√2 - 2)/(4 + 3√2) = [(2√3 + √6 - 2√2 - 2)(4 - 3√2)]/[(4 + 3√2)(4 - 3√2)] = (2√3 - 2√6 - 2√2 + 4)/(-2) = -√3 + √6 + √2 - 2 = √6 + √2 - √3 - 2

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