Question

高中三角函数急急急必采纳

高中三角函数急急急必采纳高三三角函数问题急急急必采纳已知sin（α+π/4）=√2 /10，α∈（π/2，π）
求cosα的值
求sin（2α- π/4）的值

Answer 1

解1由sin(a+π/4)=√2/10，a+π/4属于(3π/4,5π/4)
知cos(a+π/4)=-7√2/10
故cosa=cos(a+π/4-π/4)
=cos(a+π/4)cosπ/4+sin(a+π/4)sinπ/4
=-7√2/10×√2/2+√2/10×√2/2
=(-14+2)/20
=-3/5
(2)由cosa=-4/5，即sin2a=2sinacosa=24/25
cos2a=2cos^2a-1=7/25
sin(2a-π/4)
=sin2acosπ/4+cos2acosπ/4
=24/25×√2/2+7/25×√2/2
=31√2/50