given (sin²θ ) / (2+7cos²θ) = 1/9
where 3π/2< θ< 2π
find the valueog tanθ
amath
2007-11-24 5:36 am
回答 (2)
2007-11-24 5:48 am
✔ 最佳答案
(sin²θ ) / (2+7cos²θ) = 1/9 (1-cos^2 θ)/(2+7cos^2 θ) = 1/9
9-9cos^2 θ = 2+7cos^2 θ
16cos^2 θ = 7
cos^2 θ = 7/16
cos θ = sqrt(7)/4
tan θ = -sqrt(16-7)/sqrt(7) [Note that 3π/2< θ< 2π, tanθ <0]
tan θ = -3/sqrt(7) = -3sqrt(7)/7
2007-11-24 5:59 am
(sin²θ ) / (2+7cos²θ) = 1/9
9sin²θ =2+7cos²θ
9sin²θ=2(sin²θ+cos²θ)+7cos²θ
7sin²θ=9cos²θ
sin²θ/cos²θ=9/7
你明啦
tanθ=+/-1.285714
where 3π/2< θ< 2π
tanθ=-1.285714
有d野打唔到出來
當見到有integer 同sin cos 時
用sin²θ +cos²θ=1 折個integer 好好用的
2007-11-23 22:01:00 補充：
咁錯機tanθ=-1.13389
9sin²θ =2+7cos²θ
9sin²θ=2(sin²θ+cos²θ)+7cos²θ
7sin²θ=9cos²θ
sin²θ/cos²θ=9/7
你明啦
tanθ=+/-1.285714
where 3π/2< θ< 2π
tanθ=-1.285714
有d野打唔到出來
當見到有integer 同sin cos 時
用sin²θ +cos²θ=1 折個integer 好好用的
2007-11-23 22:01:00 補充：
咁錯機tanθ=-1.13389
收錄日期: 2021-04-13 18:12:13
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