2道高一三角 1 函数y=3sin(x+10°)+5sin(x+70°)的最大值是?2 在△ABC中,三边的长度为跟号a

3个回答

  • 1) 3sin(x+10°)+5sin(x+70°)

    =3sinx*cos(10°)+3cosx*sin(10°)+5sinx*cos(70°)+5cosx*sin(70°)

    =sinx*(3cos(10°)+5cos(70°))+cosx*(3sin(10°)+5sin(70°))

    而asinx+bcox的最大值为√(a^2+b^2),所以3sin(x+10°)+5sin(x+70°)的最大值为

    √((3cos(10°)+5cos(70°))^2+(3sin(10°)+5sin(70°))^2)

    =√(34+30( cos(10°)cos(70°)+sin(10°)sin(70°) )

    =√(34+30( cos(10°-70°) )

    =√(34+30( cos(-60°) )

    =√(34+30( cos(60°) )

    =√(34+30*1/2)

    =√49

    =7

    2、直角三角形,选B.显然b+c=a意味着有两边的长度平方和等于第三边的平方.