09/10/13 21:14:11
>>535 出題元の解答は…
〔補題〕
|a･cos(x) + b･sin(x)| ≦ √(a^2 + b^2),
(略証)
{a･cos(x) + b･sin(x)}^2 = a^2 + b^2 - {b･cos(x) - a･sin(x)}^2 ≦ a^2 + b^2, (終)
(1)
(与式) = (31/2) +6sin(2x) +(1/2)cos(2x) ≦ (31/2) + √{6^2 + (1/2)^2},
(2)
(与式) = 3sinθ(5+4sinθ) + 16(cosθ)^2
= 3sinθ(5+4cosθ) + 25 - (5-4cosθ)(5+4cosθ)
= 25 - (5 -3sinθ -4cosθ)(5+4cosθ)
≦ 25,
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