解析等式(sinα+cosα)/(sinα"-" cosα)=1/2左边分子、分母同时除以cos α(显然cos α≠0),得(tanα+1)/(tanα"-" 1)=1/2,解得tan α=-3,

　　∴tan 2α=2tanα/(1"-" tan^2 α)=3/4.

答案B

7.已知sin(π/4 "-" x)=3/5,则sin 2x=　　　　　.

答案7/25

8.定义运算a􀱇b=a2-ab-b2,则sinπ/6􀱇cosπ/6=　　　　　.

解析原式=sin2π/6-sinπ/6·cosπ/6-cos2π/6=-cosπ/3-1/2sinπ/3=-1/2-√3/4.

答案-1/2-√3/4

9.求下列各式的值:

(1)(2cos^2 α"-" 1)/(2tan(π/4 "-" α)sin^2 (π/4+α) );

(2)2√3tan 15°+tan215°;

(3)sin 10°sin 30°sin 50°sin 70°.

解(1)原式=cos2α/(2tan(π/4 "-" α)cos^2 (π/2 "-" π/4 "-" α) )

　　=cos2α/(2tan(π/4 "-" α)cos^2 (π/4 "-" α) )

　　=cos2α/2sin(π/4 "-" α)cos(π/4 "-" α)

　　=cos2α/sin(2×π/4 "-" 2α)

　　=cos2α/cos2α=1.

　　(2)原式=√3tan 30°(1-tan215°)+tan215°

　　=√3×√3/3(1-tan215°)+tan215°=1.

　　(3)(方法一)sin 10°sin 30°sin 50°sin 70°

　　=1/2cos 20°cos 40°cos 80°=(2sin20"°" cos20"°" cos40"°" cos80"°" )/4sin20"°" =(sin40"°" cos40"°" cos80"°" )/4sin20"°" =(sin80"°" cos80"°" )/8sin20"°" =1/16·sin160"°" /sin20"°" =1/16.

　　(方法二)令x=sin 10°sin 50°sin 70°,

　　y=cos 10°cos 50°cos 70°.

　　则xy=sin 10°cos 10°sin 50°cos 50°sin 70°cos 70°

　　=1/2sin 20°·1/2sin 100°·1/2sin 140°

　　=1/8sin 20°sin 80°sin 40°

　　=1/8cos 10°cos 50°cos 70°

　　=1/8y.

　　∵y≠0,∴x=1/8.

　　从而有sin 10°sin 30°sin 50°sin 70°=1/16.

10.导学号93774097已知函数f(x)=2cos x(sin x-cos x),x∈R.

(1)求函数f(x)图像的对称中心;

(2)求函数f(x)在区间[π/8 "," 3π/4]上的最小值和最大值.

解(1)f(x)=2cos x(sin x-cos x)=sin 2x-cos 2x-1=√2sin(2x"-" π/4)-1.

　　令2x-π/4=kπ,k∈Z,得x=kπ/2+π/8,k∈Z,

　　因此,函数f(x)的图像的对称中心为(kπ/2+π/8 ",-" 1),k∈Z.

(2)因为f(x)=√2sin(2x"-" π/4)-1在区间[π/8 "," 3π/8]上为增函数,在区间[3π/8 "," 3π/4]上为减函数,