高一数学题已知cos[(π/4)
看起来很麻烦的,可能算错了
解:cos[(π/4)+x]=3/5且
∵17π/12< x<21π/12
∴5π/3<π/4+x<2π且cos[(π/4)+x]=3/5
∴sin[(π/4)+x]=-4/5
cosx=cos[(π/4+x)-π/4]
=cos[(π/4)+x]cosπ/4+sin[(π/4)+x]sinπ/4
=(√2/2)(3/5-4/5)
=-√2/10
∴sinx=-7√2/10
[sin2x+2(sinx^2)]/(1-tgx)
=[2sinxcosx+2(sinx^2)]/(1-tgx)
=(7/25+49/25)/(1-7)
=-28/75。 全部
看起来很麻烦的,可能算错了
解:cos[(π/4)+x]=3/5且
∵17π/12< x<21π/12
∴5π/3<π/4+x<2π且cos[(π/4)+x]=3/5
∴sin[(π/4)+x]=-4/5
cosx=cos[(π/4+x)-π/4]
=cos[(π/4)+x]cosπ/4+sin[(π/4)+x]sinπ/4
=(√2/2)(3/5-4/5)
=-√2/10
∴sinx=-7√2/10
[sin2x+2(sinx^2)]/(1-tgx)
=[2sinxcosx+2(sinx^2)]/(1-tgx)
=(7/25+49/25)/(1-7)
=-28/75。
收起