化简sin^4α+cos^4α+2sin^2α*cos^2α-1
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化简sin^4α+cos^4α+2sin^2α*cos^2α-1化简sin^4α+cos^4α+2sin^2α*cos^2α-1化简sin^4α+cos^4α+2sin^2α*cos^2α-1sin^
化简sin^4α+cos^4α+2sin^2α*cos^2α-1
化简sin^4α+cos^4α+2sin^2α*cos^2α-1
化简sin^4α+cos^4α+2sin^2α*cos^2α-1
sin^4α+cos^4α+2sin^2α*cos^2α-1
=(sin²a+cos²a)²-1²
=1-1
=0
sin^4α+cos^4α+2sin^2α*cos^2α-1=[sin^2α+cos^2α]^2-1=0
化简sin^4α+cos^4α+2sin^2α*cos^2α-1
化简sin^4α+cos^2α+sin^2α*cos^2α
化简:cos^4α-sin^4α+2sin^2α化简:cos^4α-sin^4α+2sin^2α
sinα-cosα×tanα-sin^4α-sin^2α×cos^2-cos^2α 求最后结果
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已知sinΘ+cosΘ=2sinα,sinΘ*cosΘ=sin²β,求证:4cos²2α=cos²2β
已知sinθ+cosθ=2sinα,sinθ*cosθ=sin²β,求证:4cos²2α=cos²2β
化简sin(π/4+α)cosα-sin(π/4-α)sinα
化简sin(α-5π)/cos(3π-α)×cos(π/2-α)/sin(α-3π)×cos(8π-α)/sin(-α-4π)
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sin^6а+cos^6а+3sin²аcos²а化简=(sin^2α+cos^2α)(sin^4α-sin^2α*cos^2α+cos^4α)+3sin^2α*cos^2α这一步的sin四次方是怎么出来的?看不懂,求这一步的详写.
化简sin^2α+sin^2β+2sinαsinβcos(α+β)
化简:2cos²α-1/【2sin(π/4-α)sin²(π/4+α)】
化简:2cos²α-1/【2sin(π/4-α)sin²(π/4+α)】
化简sin^4α+cos^4α+1/2sin²2α=?
化简4(sinα)^2+2sinαcosα-1
2sinα-3cosα/4sinα-9cosα=-1,则9sin^2α-3sinαcosα-5=
化简4sin²α-sinαcosα-3cos²α=0