已知向量a=(2,2,1),则a的方向余弦为(). A: cosα=2/3,cosβ=2/3,cosγ=1/3 B: cosα=2/5,cosβ=2/5,cosγ=1/5
已知向量a=(2,2,1),则a的方向余弦为(). A: cosα=2/3,cosβ=2/3,cosγ=1/3 B: cosα=2/5,cosβ=2/5,cosγ=1/5
将数学表达式Cos A: Cos(a+b+5*exp(2) B: Cos (a+ C: +5*exp(2 D: Cos(a+b +5*ln(2) E: Cos (a+ F: +5*ln(2)
将数学表达式Cos A: Cos(a+b+5*exp(2) B: Cos (a+ C: +5*exp(2 D: Cos(a+b +5*ln(2) E: Cos (a+ F: +5*ln(2)
\(\int { { {\sin }^{2}}x { { \cos }^{5}}xdx}\)=( ) A: \(\frac{1}{3} { { \sin }^{3}}x-\frac{2}{5} { { \sin }^{5}}x+\frac{1}{7} { { \sin }^{7}}x+C\) B: \(\frac{2}{3} { { \sin }^{3}}x-\frac{1}{5} { { \sin }^{5}}x-\frac{1}{7} { { \sin }^{7}}x+C\) C: \(\frac{1}{3} { { \cos }^{3}}x-\frac{2}{5} { { \cos }^{5}}x+\frac{1}{7} { { \cos }^{7}}x+C\) D: \(\frac{2}{3} { { \cos }^{3}}x-\frac{1}{5} { { \cos }^{5}}x-\frac{1}{7} { { \cos }^{7}}x+C\)
\(\int { { {\sin }^{2}}x { { \cos }^{5}}xdx}\)=( ) A: \(\frac{1}{3} { { \sin }^{3}}x-\frac{2}{5} { { \sin }^{5}}x+\frac{1}{7} { { \sin }^{7}}x+C\) B: \(\frac{2}{3} { { \sin }^{3}}x-\frac{1}{5} { { \sin }^{5}}x-\frac{1}{7} { { \sin }^{7}}x+C\) C: \(\frac{1}{3} { { \cos }^{3}}x-\frac{2}{5} { { \cos }^{5}}x+\frac{1}{7} { { \cos }^{7}}x+C\) D: \(\frac{2}{3} { { \cos }^{3}}x-\frac{1}{5} { { \cos }^{5}}x-\frac{1}{7} { { \cos }^{7}}x+C\)
( )sin(5x+2)'=cos(5x+2) A: 5 B: 1/5 C: 2 D: 1/2
( )sin(5x+2)'=cos(5x+2) A: 5 B: 1/5 C: 2 D: 1/2
3.将正弦电压 u = 10 cos ( 314t+30( ) V 施加于感抗 XL = 5 ( 的电感元件上,则通过该元件的 电流i = ( )。 A: A、2 cos( 314t-60() A B: 2 cos( 314t+60() A C: 2 cos( 314t+30() A D: 2 cos( 314t-30() A
3.将正弦电压 u = 10 cos ( 314t+30( ) V 施加于感抗 XL = 5 ( 的电感元件上,则通过该元件的 电流i = ( )。 A: A、2 cos( 314t-60() A B: 2 cos( 314t+60() A C: 2 cos( 314t+30() A D: 2 cos( 314t-30() A
已知tan(π+α)=2,则cos²α等于y A: 4/5 B: 3/5 C: 2/5 D: 1/5
已知tan(π+α)=2,则cos²α等于y A: 4/5 B: 3/5 C: 2/5 D: 1/5
( )[sin(5x+2)]'=cos(5x+2). A: 5 B: 1/5 C: 2 D: 1/2
( )[sin(5x+2)]'=cos(5x+2). A: 5 B: 1/5 C: 2 D: 1/2
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))
$\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$
$\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$
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