原式=sinα/cosα *sinα*cosα
=sin^2α
化簡:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-α)
化簡:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-α)
RT.
RT.
數(shù)學人氣:876 ℃時間:2020-03-20 19:10:50
優(yōu)質解答
我來回答
類似推薦
- 化簡:cos(a-π/2)/sin(5π/2+a)*sin(a-2π)*cos(2π-a)
- 化簡sin(θ-5π)cos(π/2+θ)cos(4π-θ)/cos(3π-θ)sin(θ-3π)sin(-θ-4π)
- (sinα+cosα)^2化簡
- 化簡cos(α-π/2)/sin(α+5π/2)*sin(α-2π)*cos(2π-α),因為我剛學,所以要具體點的過程哦~
- [sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化簡
- Complete this passage with some of the words a
- 1.小張該月支付的平段、谷段電價每千瓦時各多少元?2.如不使用分時電價結算,5月份小張家多支付電費多少元
- are you from canada,too
- 已知⊙O半徑為6,有一條弦AB長63,則AB所對的圓周角為( ?。?A.30° B.60° C.30°或60° D.60°或120°
- 已知等腰△ABC中,AB=AC,D是BC邊上一點,連接AD,若△ACD和△ABD都是等腰三角形,則∠C的度數(shù)是_.
- we can seed some ants and butterflias.(改為否定回答)
- 驗證機械能守恒定律 g怎么處理