简单数学1.(1)已知tanX=
1。(1)已知tanX=-1/3,求(5cosX-sinX)/(sinX+2cosX)的值;
(2)求证:sin2X/(1+sinX+cosX)=(sinX+cosX-1)
(1)分子分母同除cosX得到
(5cosX-sinX)/(sinX+2cosX)
=(5-tanX)/(tanX+2)=(16/3)/(5/3)=16/5。
(2)(1+sinX+cosX)(sinX+cosX-1)
=[(sinX+cosX)+1][(sinX+cosX)-1]
=(sinX+cosX)^2-1=(sinX)^2+(cosX)^2+2sinXcosX-1
=2sinXcosX=sin2X。
...全部
1。(1)已知tanX=-1/3,求(5cosX-sinX)/(sinX+2cosX)的值;
(2)求证:sin2X/(1+sinX+cosX)=(sinX+cosX-1)
(1)分子分母同除cosX得到
(5cosX-sinX)/(sinX+2cosX)
=(5-tanX)/(tanX+2)=(16/3)/(5/3)=16/5。
(2)(1+sinX+cosX)(sinX+cosX-1)
=[(sinX+cosX)+1][(sinX+cosX)-1]
=(sinX+cosX)^2-1=(sinX)^2+(cosX)^2+2sinXcosX-1
=2sinXcosX=sin2X。
所以sin2X/(1+sinX+cosX)=(sinX+cosX-1)。
2。(1)已知0<X<π/4,sin(π/4-X)=5/13,求
cos2X/cos(π/4+X)的值;
因为sin(π/4-X)=5/13,所以cos(π/4-X)=12/13,
sin(π/4+X)=cos(π/4-X)=12/13,
cos2X/cos(π/4+X)=sin(π/2+2X)/cos(π/4+X)=
=sin[2(π/4+X)]/cos(π/4+X)
=2sin(π/4+X)cos(π/4+X))/cos(π/4+X)
=2sin(π/4+X)=24/13
(2)已知A+B=π/4,求证:(1+tanA)×(1+tanB)=2。
1=tan(π/4)=tan(A+B)=(tanA+tanB)/(1-tanAtanB),
tanA+tanB=1-tanAtanB,
1+tanA+tanB+tanAtanB=2,
所以(1+tanA)×(1+tanB)=2。
(3)求值:㏒2cosπ/9+㏒2cos2π/9+㏒2cos4π/9。
利用sin8π/9=sinπ/9
㏒2[(cosπ/9)(cos2π/9)(cos4π/9)]
=㏒2[(sinπ/9)(cosπ/9)(cos2π/9)(cos4π/9)/(sinπ/9)]
=㏒2[(sin2π/9)(cos2π/9)(cos4π/9)/(2sinπ/9)]
=㏒2[(sin4π/9)(cos4π/9)/(4sinπ/9)]
=㏒2[(sin8π/9)/(8sinπ/9)]
=㏒2(1/8)
=-3。
。收起