Question

三角函数，求大神帮忙。。

Answer 1

 (1)第一题是图片

（2）不妨设A≥π/3,则B+C≤π-π/3=2π/3.
sin3A+sin3B+sin3C
=sin3A+2sin[3/2*(B+C)]cos[3/2*(B-C)]
≤sin3A+2sin[3/2*(B+C)].
记α=3/2*(B+C),则0≤α≤π,
且A=π-(B+C)=π-(2/3)α.
于是,
sin3A+sin3B+sin3C
≤sin(3π-α)+2sinα
=sin2α+2sinα
=2sinα(1+cosα)
=8sin(α/2)[cos(α/2)]^3
=8根[1/3*3(sin(α/2))^2(cos(α/2))^6]
≤8根{1/3*1/4^4*[(3sin(α/2))^2+(cos(α/2))^2+(cos(α/2))^2+(cos(α/2))^2]^4}
=(3根3)/2
=根(27/4)
∴sin3A+sin3B+sin3C≤根(27/4),
当且仅当A=7π/9,B=C=π/9时,等号成立.

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