已知sinx=-3/5,x∈(3π/2,2π)求sin2x和tan(π-2x)的值.有步驟線上等.

已知sinx=-3/5,x∈(3π/2,2π)求sin2x和tan(π-2x)的值.有步驟線上等.


解:由sinx=-3/5,x∈(3π/2,2π)可得cosx=4/5
sin2x=2sinxcosx=-24/25 tanx=sinx/cosx=-3/4
所以tan(π-2x)=-(2tanx)/[1-(tanx)^2]=24/7



已知-pi/3小於等於x小於等於pi/4,fx=tan^2x+2tanx+2,求fx的最值及相應
x值


最小值是1,此時x取-pi/4.最大值是5,此時x取pi/4



sin(Pai/12)-根號3*cos(Pai/12)等於多少,怎麼算


sin(π/12)-√3*cos(π/12)=2*[(1/2)*sin(π/12)-(√3/2)*cos(π/12)] =2*sin(π/12-π/3)=2sin(-π/4)=-2sin(π/4)=-2√2