已知α为锐角,且cos（α＋π/4）＝3/5,则sinα＝
问题描述：
已知α为锐角,且cos（α＋π/4）＝3/5,则sinα＝
答
sin(a+兀/4)=4/5
原式变为cos(a+兀/4-兀/4)
cos(a+兀/4)cos(兀/4)+sin(a+兀/4)sin(兀/4)=cosa
答
cos(α＋π/4）＝3/5
sin(α＋π/4）＝4/5
sinα=sin[(a+π/4)-π/4]=(√2/2)sin(a+π/4)-(2/2)cos(a+π/4)
=(√2/2)(4/5-3/5)=√2/10
答
解
∵a是锐角
即a∈(0.π/2)
∴a+π/4∈(π/4,3π/4)
∴sin(a+π/4)>0
∵cos(a+π/4)=3/5——利用sin²a+cos²a=1
∴sin(a+π/4)=4/5
∴sina
=sin[(a+π/4)-π/4]
=sin(a+π/4)cosπ/4-cos(a+π/4)sinπ/4
=√2/2(4/5-3/5)
=√2/10