osπ/4cosx-sinπ/4sinx=3/5 cosx-sinx=3√2/5 平方 cosx+sinx-2sinxcosx=18/25 1-2sinxcosx=18/25 sinxcosx=7/50 sinx+cosx=√2sin(x+π/4) 5π/4
若cos(π/4+x)=3/5,17π/12<x <7π/4,求(sin2x+2cosx)/1-tanx的值
1个回答
相关问题
-
若cos(π/4+x)=3/5,17/12π<x<7/4π,求(sin2x+2sinx)/(1-tanx)的值
-
已知cos(π/4+x)=3/5,17π/12<x<7π/4,求sin2x+2sin平方x/(1-tanx)的值
-
设cos﹙x+[π/4]﹚=[3/4],[17π/12]<x<[7π/4],求cos2x•[1-tanx/1+tanx]
-
已知cos(π/4+x)=3/5,x∈(3π/4,7π/4),求sin2x+2Sin^x /1-tanx的值
-
已知cos(x+π/4)=3/5,5π/4<x<7π/4,求 sin2x+2sin²x ÷ 1-tanx 的值
-
已知cos(π/4+x)=3/5,5/4π<x<7π/4,求(sin2x+2sin²x)/(1-tanx)
-
已知cos(π/4+x)=4/5,x∈(-π/2,-π/4).求(sin2x-2sin²x)/(1+tanx)
-
已tanx=4/3,求cos(2x-π/3)cos(π/3-x)-sin(2x-π/3)sin(π/3-x)
-
sinx=-15/17 x属于(3π/2,2 π) 求sin(x+π/4) cos(π/4- x)的值
-
已知cos(x+45º)=3/5且17π/12<x<7π/4,求sin2x+2sin²x/1-tan