Question

tanx=1/7,siny=根号10/10,则x+2y=?3Q

Answer 1

siny=√10/10,∴cosy=3√10/10 则sin2y=2sinycosy=3/5,∴cos2y=4/5 tanx=sinx/cosx=1/7,即7sinx=cosx.而sinx+cos=1,即sinx+(7sinx)=1 解得sinx=1/√50=√2/10,∴cosx=7√2/10 有sin(x+2y)=sinxcos2y+sin2ycosx=4√2/10*5+3*7√2/10*5=√2/2 则x+2y=π/4+2kπ（k属于正整数）