Question

帮忙求下三角函数的值域求函数y=cos^2x-sinx的值域.特别是求值域的那几步

Answer 1

由y=cos^2x-sinx

y=1-sin^2x-sinx

=-(sin^2x+2*(1/2)*sinx+1/4)+5/4

=-(sinx+(1/2))^2+(5/4)

而:-1≤sinx≤1

-1/2≤sinx+(1/2)≤3/2

所以：0≤|sinx+(1/2)|≤3/2

所以：-(3/2)^2≤-(sinx+(1/2))^2)≤0

所以：-(3/2)^2+(5/4)≤-(sinx+(1/2))^2)+(5/4)≤5/4

即：-1≤y≤5/4