若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)

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若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)
若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)

若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)
sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)式子有误.是不是:
[sin(-α)+(sin-90°-α)]/[cos(540°-α)+cos(-270°-α)]

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若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α) 若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+cos(-270º-α)】=? sin(10°)+sin(50°)-sin(70°) =sin(10°)+sin(50°)-sin(70°) = 已知sinα=-0.4,求下列三角函数值 sin(-α) sin(3π-α) sin(α-π) sin(α-10π) 证明sin(α+β)sin(α-β)=sinα-sinβ 计算:sin方10°+sin方20°+sin方30°+sin方40°+sin方50°+sin方60°+sin方70°+sin方80°+sin方90°= 诱导公式的推导,如何利用公式sin(π+α)=-sinα 和sin(-α)=-sinα 得到sin(π-α)=sinα ? 已知8sinα+10cosβ=5,8cosα+10sinβ=5√3,求证:sin(α+β)=-sin(π/3+α) 计算:sin²10°+sin²20°+sin²30°+.sin²80°= 计算:sin²10°+sin²20°+sin²30°+……sin²80°= sin²10°+sin²20°+sin²30°+……+sin²80°=? 若α为锐角,化简√(1-2sinα+sin²α)= 若sin(π+α)=1/√10,则sec(-α)+sin(-α-90º)/csc(540º-α)-cos(-α-270º)的值 sin²10°+sin²20°+sin²30°+……+sin²80°+sin²90°=? 1 已知sin(90°+α)sin(90°-α)=1/6,且α∈(90°,180°),求sin4α的值2 已知sin(90°-α)=(7√2)/10,cos2α=7/25,求sinα及tan(α+60°)的值> 求值:(tan10°-√3)×cos10°/sin50° 求证:[sin(2α+β)]/sinα- 2cos(α+β)=sinβ/cosα 化简cosθ+cos(θ+2π/3)+cos(θ+4π/3) 证明:sin(α+β)sin(α-β)=sin^2α-sin^2β 若sin^2β-sin^2α=m,则sin(α+β)sin(α-β) 1、sinαsinβ-cosαcosβ=2、化简cos(α-β)cos(α+β)+sin(α-β)sin(α+β)=3、若α,β都是锐角,且sinα=1/根号10,cosβ=2/根号5,则cos(α+β)=4、cos75°-cos15°=5、在三角形ABC中,cosAcosB>sinAsinB,则此三角形是什么三角