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若tAnα=2,则sin2α=?

sinα=-3cosα ∵sin2α+cos2α=1 ∴(-3cosα)2+cos2α=1 ∴10cos2α=1 ∴cos2α=1/10 ∴2sinαcosα=-6cos2α=-3/5

sin²α-2sinαcosα-cos²α =(sin²α-2sinαcosα-cos²α)/(sin²α+cos²α) =(tan²α-2tanα-1)/(tan²α+1) 分子、分母同除以cos²α =-1/5

∵tanα=2,∴sin2α=2sinαcosα=2sinαcosαcos2α+ sin2α=2tanα1+tan2α=45,故答案为 45.

cos2asin2a=(cosαsinα)2=(1tanα)2=14.故答案为:14.

∵tanα=3,sin2αcos2α=2sinα?cosαcos2α=2tanα=6,故答案为:6.

∵tanθ=1,则sin2θ-2cos2θ=2sinθcosθ-2cos2θ=2sinθcosθ?2 cos2θcos2θ+ sin2θ=2tanθ?21+tan2θ=0,故答案为 0.

解由sin2θ=-4/5,且θ是第二象限角, 知2θ是第三象限角 故cos2θ=-3/5 故tanθ =sinθ/cosθ =2cosθsinθ/2cos^2θ =sin2θ/(1+cos2θ) =(-4/5)/(1+(-3/5)) =(-4/5)/(2/5) =-2

cos²α+2sin2α =cos²α+4sinαcosα =cos²α(1+4tanα) =4cos²α/(cos²α+sin²α) =4/(1+tan²α) =64/25

∵sin2α+cos2α=1∴2sin2θ-3sinθcosθ=2sin2θ?3sinθcosθsin2θ+cos2θ=2tan2θ?3tanθtan2θ+1=2×22?3×222+1=25,故答案为:25.

原式=(sin²α+2sinαcosα-3cos²α)/1 =(sin²α+2sinαcosα-3cos²α)/(sin²α+cos²α) 上下除以cos²α =(tan²α+2tanα-3)/(tan²α+1) =1

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