搜搜做题作业网已知cos x=-12/13 ,x∈(π,3π/2),求tan(x-π/4)的值.
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已知cos x=-12/13 ,x∈(π,3π/2),求tan(x-π/4)的值.
额,郁闷,似乎很简单,但是我不会,麻烦给点详细步骤```
额,郁闷,似乎很简单,但是我不会,麻烦给点详细步骤```
因为 x∈(π,3π/2)
所以 sinx < 0
sinx = -√[1 - (cosx)^2] = -√[1 - (-12/13)^2] = -√(1 - 144/169) = -√25/169
= - 5/13
tan(x-π/4)
= sin(x-π/4)/cos(x-π/4)
= [sinx cos(π/4) - cosxsin(π/4)]/[cosxcos(π/4) + sinxsin(π/4)]
= (sinx - cosx)/(cosx + sinx)
= [-5/13 - (-12/13)]/(-12/13 - 5/13)
= - 7/17
所以 sinx < 0
sinx = -√[1 - (cosx)^2] = -√[1 - (-12/13)^2] = -√(1 - 144/169) = -√25/169
= - 5/13
tan(x-π/4)
= sin(x-π/4)/cos(x-π/4)
= [sinx cos(π/4) - cosxsin(π/4)]/[cosxcos(π/4) + sinxsin(π/4)]
= (sinx - cosx)/(cosx + sinx)
= [-5/13 - (-12/13)]/(-12/13 - 5/13)
= - 7/17
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