两道三角函数的高一题
1.sinx-cosx=负跟号5/2,则tanx+1/tanx的值2.(1-cosx)(1/sinx+1/tanx)=...
1.sin x-cos x=负跟号5/2,则tan x+1/tan x的值 2.(1-cos x)(1/sin x+1/tan x)=
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⒈∵sin x-cos x=-(√5)/2
∴(sin x-cos x)^2=5/4
(sin x)^2-2(sin x)(cos x)+(cos x)^2=5/4
1-2(sin x)(cos x)=5/4
∴(sin x)(cos x)=-(1/8)
∴tan x+1/tan x
=(sin x)/(cos x)+(cos x)/(sin x)
=[(sin x)^2+(cos x)^2]/(sin x)(cos x)
=1/(sin x)(cos x)
=1/-(1/8)
=-8
⒉(1-cos x)(1/sin x+1/tan x)
=(1-cos x)(1/sin x+cos x/sin x)
=(1-cos x)[(1+cos x)/sin x]
=[1-(cos x)^2]/sin x
=(sin x)^2/sin x
=|sin x|
∴(sin x-cos x)^2=5/4
(sin x)^2-2(sin x)(cos x)+(cos x)^2=5/4
1-2(sin x)(cos x)=5/4
∴(sin x)(cos x)=-(1/8)
∴tan x+1/tan x
=(sin x)/(cos x)+(cos x)/(sin x)
=[(sin x)^2+(cos x)^2]/(sin x)(cos x)
=1/(sin x)(cos x)
=1/-(1/8)
=-8
⒉(1-cos x)(1/sin x+1/tan x)
=(1-cos x)(1/sin x+cos x/sin x)
=(1-cos x)[(1+cos x)/sin x]
=[1-(cos x)^2]/sin x
=(sin x)^2/sin x
=|sin x|
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