標題:

Trigonometry F.4 (not A. maths)

發問:

1. cos^2 θ = sinθ(1-sinθ) [solve it, and for 0°≧θ≧360°] 2. 1/4+3cosx [solve it, and for 0°≧θ≧360°] 3. tan^2 θ - ( √3 - 1 ) tanθ = √3 [solve it, and for 0°≧θ≧360°,corr.to nearst degree.] (U may explain how to do, or just write the step~I really don't know...") Thanks so much~

最佳解答:

1.(cosθ)^2=sinθ(1-sinθ) (cosθ)^2=sinθ-(sinθ)^2 (cosθ)^2+(sinθ)^2=sinθ 1=sinθ θ=90° 2.cannot be solved because of different variables and not an (in)equality, we can only get a range. Assume x=θ, since -1 =< cosx =< 1, -2.75 =< 1/4+3cosx =< 3.25, where =< means less than or equal to. 3.(tanθ)^2-(√3 - 1) tanθ = √3 Let y=tanθ y^2-(√3 - 1)y-√3=0 solve the quadratic equation: y=√3 or y=-1 tanθ=√3 or tanθ=-1 θ=60°or θ=240°or θ=135°or θ=315°

 

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其他解答:

Detailed!|||||1. cos^2x=sinx(1-sinx) cos^2x=sinx-sin^2x cos^2x+sin^2x=sinx 1=sinx x=90 2.條式=乜？ 3.tan^2 x - ( √3 - 1 ) tanx = √3 tan^2 x - ( √3 - 1 ) tanx- √3= 0 By quadratic formula,tanx={( √3 - 1 )+-√[( √3 - 1 )^2-4√3]}/2 =......8758B59A7FA1EEA7
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