已知arcsinx+arctg(1/7)=π/4,则x=?要有过程！
热心网友
令a=arctg(1/7)---tga=1/7.(0cosa=1/seca=1/√[1+(tga)^2]=1/√(1+1/49)=1/(√50/7)=7√2/10---sina=cosatga=7√2/10*1/7=√2/ csinx+arctg(1/7)=π/4---arcsinx=π/4-a---x=sin(π/4-a)=sin(π/4)cosa-cos(π/4)sina=√2/2*(cosa-sina)=√2/2(7√2/10-√2/10)=√2/2*6√2/10=3/5.
热心网友
arcsinx=π/4-arctg(1/7)x=sin[π/4-arctg(1/7)]=sinπ/4*cos[arctg(1/7)]-cosπ/4*sin[arctg(1/7)]=1/genhao2*7/genhao50-1/genhao2*1/genhao50=6/genhao100=6/10=3/5注:arctg(1/7)=a,tga=1/7,sina=1/genhao50,cosa=7/genhao50