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標題:
A Maths問題
It is given that tan x=-9/40 and 3π/2
最佳解答:
sinx/cosx=-9/40, sin^2x+cos^2x=1, so, cosx = 40/41 and sinx = -9/41 as 3π/2<2π. a.sin(3π/2+x) =sin(2π-(π/2-x)) =-sin(π/2-x) =-cosx =-40/41 b.cos(π+x) = -cosx =-40/41 c.sin(x+13π) = sin(x+π) =-sinx =9/41 d.tan(7π/2-x) =tan(3π/2-x) =tan(2π-(π/2+x)) =-tan(π/2+x) =1/tanx =-40/9
其他解答:
A Maths問題
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發問:It is given that tan x=-9/40 and 3π/2
最佳解答:
sinx/cosx=-9/40, sin^2x+cos^2x=1, so, cosx = 40/41 and sinx = -9/41 as 3π/2<2π. a.sin(3π/2+x) =sin(2π-(π/2-x)) =-sin(π/2-x) =-cosx =-40/41 b.cos(π+x) = -cosx =-40/41 c.sin(x+13π) = sin(x+π) =-sinx =9/41 d.tan(7π/2-x) =tan(3π/2-x) =tan(2π-(π/2+x)) =-tan(π/2+x) =1/tanx =-40/9
其他解答:
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