复数z=6i的三角形式为（） A: cosπ/2+isinπ/2 B: cosπ/2-sinπ/2 C: sinπ/2+icosπ/2 D: sinπ/2-icosπ/2

cos2x为什么等于cos^2-sin^2

函数[img=79x27]180355ae2690a03.png[/img]在x=2处的二阶泰勒展开式为 A: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 B: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 C: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2 D: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2

用五点法作出函数y=2-sin(x-π/2),x属于π/2,5π/2的图像

sin(α-β)cosβ+cos(α-β)sinβ=( ) A: sin(α-2β) B: cos(α-2β) C: sinα D: cosα

利用三角函数来化简符号表达式http://edu-image...cos(x)^2-sin(x)^2;F=

以下不能正确计算代数式值的C语言表达式是（） A: 1/3*sin(1/2)*sin(1/2) B: sin(0.5)*sin(0.5)/3 C: pow(sin(0.5),2)/3 D: 1/3.0*pow(sin(1.0/2),2)

曲线积分$$\int_{(0,0}^{(x,y)}(2x\cos y-y^2\sin x)dx+(2y\cos x-x^2\sin y)dy=$$ A: $y^2\cos x+x^2\cos y$ B: $x^2\cos x+y^2\cos y$ C: $x^2\sin y+y^2\sin x$ D: $x^2\sin x+y^2\sin y$

已知sin（α+β）=2/1,sin（α-β）=3/1求证：sinαcosβ=5cosαsinβ

$\int {{{\sin 2x} \over {1 + {{\sin }^4}x}}} {\rm{d}}x = $ A: $\arctan (\sin x) + C$ B: $\arctan ({\sin ^2}x) + C$ C: ${\arctan ^2}(\sin x) + C$ D: $ - {\arctan ^2}(\sin x) + C$