若arcsinx+arctg1/7=π/4，则x=___?过程 谢谢～～～
热心网友
arcsinx=π/4-arctan1/7 x=sin(arcsinx)=sin(π/4-arctan1/7)=sinπ/4cos(arctan1/7)-cosπ/4sin(arctan1/7)cos(arctan1/7)=(1/(1+tan(arctan1/7)^2)^0.5=(49/50)^0.5sin(arctan1/7)=(1/50)^0.5x=3/5
热心网友
令tgA =1/7 （#）p表圆周率sin(p/4-A)=xsin(p/4)*cosA-sinA*cos(p/4)=xsin(p/4) and cos(p/4) 易得答案由（#）可做一三角形两直角边分别为本 1 和7由便可求出x=3/5
热心网友
利用：asinx+bsinx=[√(a^2+b^2)]sin[x+arctg(b/a)]有：arcsinx+arctg(1/7)=π/4arcsinx=(π/4)-arctg(1/7)sin(arcsinx)=sin[(π/4)-arctg(1/7)]既：x=sin[(π/4)-arctg(1/7)]=[7sin(π/4)-cos(π/4)]/√(7^2+1^2)=3/5
热心网友
arcsinx+arctg1/7=π/4 tg(arcsinx+arctg1/7)=tgπ/4(tg(arcsinx)+tgarctg1/7)/(1-tg(arcsinx)*tgarctg1/7)=1x/(1-x^2)^1/2+1/7=1-x/7(1-x^2)^1/2 (-1