图示结构跨中点截面C的弯矩(以下侧受拉为正)是:() A: ql/8cosα; B: -ql/8cosα; C: ql/8cosα; D: -ql/8cosα
图示结构跨中点截面C的弯矩(以下侧受拉为正)是:() A: ql/8cosα; B: -ql/8cosα; C: ql/8cosα; D: -ql/8cosα
cos(x)*cos(x/2)*cos(x/4)*cos(x/8).cos(x/(2^(n-1))
cos(x)*cos(x/2)*cos(x/4)*cos(x/8).cos(x/(2^(n-1))
将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)
将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)
已知cosα=-8/17,求sinα和tanα的值
已知cosα=-8/17,求sinα和tanα的值
已知\( {y^{(6)}} = \cos x \),则\( {y^{(8)}} = - \sin x \)( ).
已知\( {y^{(6)}} = \cos x \),则\( {y^{(8)}} = - \sin x \)( ).
序列x(n)=cos(3πn/8)的周期是()。 A: 3 B: 8 C: 16 D: 不存在
序列x(n)=cos(3πn/8)的周期是()。 A: 3 B: 8 C: 16 D: 不存在
【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
【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
【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)
【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)
\(设f(x,y,z)=\frac{x\cos y+y\cos z+z\cos x}{1+\cos x+\cos y+\cos z},则df|_{(0,0,0)}=(\,)\)
\(设f(x,y,z)=\frac{x\cos y+y\cos z+z\cos x}{1+\cos x+\cos y+\cos z},则df|_{(0,0,0)}=(\,)\)
sin(α-β)cosβ+cos(α-β)sinβ=( ) A: sin(α-2β) B: cos(α-2β) C: sinα D: cosα
sin(α-β)cosβ+cos(α-β)sinβ=( ) A: sin(α-2β) B: cos(α-2β) C: sinα D: cosα