举一反三
- 【计算题】已知sinα+cosα=1,求:(1)sinαcosα; (2)sin α-cos α; (3)sin α-cos α
- （1-sin^6α-cos^6α）/（sin^2α-sin^4α）的值
- 【单选题】sin ( α+β ) = A. sinαcosβ-cosαsinβ B. cosαsin β-sin αcos β C. sinαcosβ+cosαsinβ D. cos αcos β-sin α sin β
- 已知sin(3派+@)=1/3,求cos(派+@)/cos@[cos(派-@)-1]+cos(@-2派)/sin(@-3派/2)cos(@-派)-sin(3派/2+@)的值
- $\int \sin^3 x \cos x dx = $ A: $\frac{\sin^4 x}{4} +C$ B: ${\sin^4 x} +C$ C: $\frac{\cos^4 x}{4} +C$ D: $\frac{\cos^4 x}{4} +C$
内容
- 0
【单选题】设y=sin(cos(x)),求 结果为:(本题10.0分) A. cos(cos(x))*cos(x)+ sin(cos(x))*sin(x)^2 B. - cos(cos(x))*cos(x) - sin(cos(x))*sin(x)^2 C. - cos(cos(x))*cos(x)^2 - sin(cos(x))*sin(x)^2 D. - cos(cos(x))*cos(x) ^2- sin(cos(x))*sin(x)
- 1
已知sinα+sinβ+sinγ=0，cosα+cosβ-cosγ=0，则cos(α-β)的值是[ ]
- 2
下列计算正确的是() A: log2cos7π4＝－12 B: 若f(cos x)＝cos 2x，则f(sin 30°)＝12 C: 若sin(π＋α)＝－12，则sin(4π－α)＝－12 D: 设tan(π＋α)＝2，则sin（α－π）＋cos（π－α）sin（π＋α）－cos（π－α）＝1
- 3
求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)
- 4
sin(α-β)cosβ+cos(α-β)sinβ=( ) A: sin(α-2β) B: cos(α-2β) C: sinα D: cosα