A. $\frac{3-\sqrt{5}}{8}$
B. $\frac{-1+\sqrt{5}}{8}$
C. $\frac{3-\sqrt{5}}{4}$
D. $\frac{-1+\sqrt{5}}{4}$
若sin(α+β)+cos(α+β)=2sqrt(2)cos(α+(π)/(4))sinβ,则( )A. tan(α-β)=1B. tan(α+β)=1C. t
已知sin(α-β)=(1)/(3),cosαsinβ=(1)/(6),则cos(2α+2β)=( )A. $\frac{7}{9}$B. $\frac{1}{
已知复数=dfrac (2-2i)(1+sqrt {3)i},则=dfrac (2-2i)(1+sqrt {3)i}=dfrac (2-2i)(1+sqrt {
已知曲线方程为 ) x=sqrt (2)cos t y=sin t .已知曲线方程为,则曲线在时的法线方程为().
已知a=((31))/((32)),b=cos(1)/(4),c=4sin(1)/(4),则( )A. c>b>aB. b>a>cC. a>b>cD. a>c>
若sin2α=(sqrt(5))/(5),sin(β-α)=(sqrt(10))/(10),且α∈[(π)/(4),π],β∈[π,(3π)/(2)],则α+β
已知(sin )^2theta -(cos )^2theta =dfrac (2sqrt {2)}(3),求(sin )^2theta -(cos )^2the
若(1+sqrt(2))5=a+bsqrt(2)(a,b为有理数),则a+b=( )A. 45B. 55C. 70D. 80
下列复数① sqrt[4](6)(cos (pi)/(8)+i sin (pi)/(8));② sqrt[4](6)(cos (pi)/(8)-i sin (p
dfrac (sqrt {3)}(2)sin (15)^circ +dfrac (1)(2)cos (15)^circ 的值为( )A.dfrac