>>523 の〔類題〕
・1≦K≦√3 のとき
sin(A) + sin(B) + sin(C) > K{cos(A)+cos(B)+cos(C)} + 1 - (√2)(K-1),
・0≦K≦1のとき
sin(A) + sin(B) + sin(C) > K{cos(A)+cos(B)+cos(C)} + (2-K) + (1-K)(1/3)C,
(略証)
0≦K≦√3 と C≦π/3 より
cos(C/2) - K・sin(C/2) ≧ (√3 -K)/2 ≧ 0,
sin(A) + sin(B) > K{cos(A)+cos(B)-sin(C)} + 1 + cos(C),
sin(A) + sin(B) + sin(C) > K{cos(A)+cos(B)+cos(C)} + 1 + (1-K){sin(C)+cos(C)},
ところで、 C≦π/3 より
1 + (1/3)C ≦ cos(C) + sin(C) ≦ √2,
(終)
>>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,
http://www.casphy.com/bbs/test/read.cgi/highmath/1102511185/111-112