(sin5π 12 cos5π 12)(sin5π 12-cos5π 12)

1,cos5π/12·sinπ/12=cos(π/4+π/6)*sin(π/4-π/6)=(√6-√2)/4*(√6+√2)/4=1/42.√[cos4-sin^2(2)+2]=√[cos^2(2)-

不对,左边等于-1/12,右边等于0

原式=[sin（π2−π12）-sinπ12][sin（π2−π12）+sinπ12]=（cosπ12-sinπ12）（cosπ12+sinπ12）=cos2π12-sin2π12=cosπ6=32．

(sin5π/12+cos5π/12)(sinπ/12-cosπ/12)=[sin(π/2-π/12)+cos(π/2-π/12)](sinπ/12-cosπ/12)=(cosπ/12+sinπ/12

构造直角三角形ABCC=π/2,B=5π/12,A=π/12sinπ/12=sinA=对边/斜边=BC/ABcos5π/12=cosB=邻边/斜边=BC/AB故cos5π/12等于sinπ/12再问：

cos5π/8*cosπ/8=-cos(π-5π/8)*cosπ/8=-cos3π/8*cosπ/8=-cos(π/2-π/8)*cosπ/8=-sin(π/8)*cosπ/8=-sin(π/4)/2

证明cos5π/12=cos(π/2-5π/12)=sinπ/12∴cos²5π/12+cos²π/12=sin²π/12+cos²π/12=1（cos5π/1

cos5/12πcosπ/12+cosπ/12sinπ/6=cosπ/12（cos5/12π+sinπ/6）=cosπ/12（cos(π/2-π/12)+sinπ/6）=cosπ/12（sin(π/1

(sin5π/12-sinπ/12)(cos5π/12+cosπ/12)=(cosπ/12-sinπ/12)(sinπ/12+cosπ/12)=cos²(π/12)-sin²(π/

【注：sin2x=2sinxcosx.sin(π-x)=sinx】原式=（1/2)×2sin(5π/12)cos(5π/12)=(1/2)sin(5π/6)=(1/2)sin[π-(π/6)]=(1/

sin（1/4π+1/6π）×cos（1/4π-1/6π）然后用正弦余弦的和角差角公式算就行了

cos5分之πcos10分3π-sin5分之πsin10分之3π=cos(5分之π+10分3π)=cos2分之π=0

sinπ/14sin3π/14sin5π/14=cos6π/14cos4π/14cos2π/14=cosπ/7cos2π/7cos3π/7……(1)因为2sinθcosθ=sin2θ,又sin3π/7

令x=sinα-sin5α,y=cosα+cos5α当α→π/2,x,y均趋于0由洛必达法则可得:对x,y同时求导所以x的导数为cosa-5cos5a,y的导数为-sina-5sin5a当α→π/2,

原式=(cos10θ+i*sin10θ)/(cos9θ-i*sin9θ)=(cos10θ+i*sin10θ)/[cos(-9θ)+i*sin(-9θ)]=cos19θ+i*sin19θ.

因为sin(5派/6）=1/2,cos(5派/6）=--（根号3）/2,所以角a终边上的这一点在第四象限,因为这一点的坐标为（1/2,--（根号3）/2）,所以这一点到原点的距离为1,所以cosa=(

应该限定a是正角.因为一个角终边与单位圆的焦点坐标为(cosA,sinA)所以知道cosa=sin5π/6,sina=cos5π/6如果a是正的话就最小值就是5π/3了再问：一个角终边与单位圆的焦点坐

建立原点坐标,先根据角α终边上一点坐标为(sin5π/6,cos5π/6)计算出原点在该坐标的距离为1那么sinα=对边/斜边=(cos5π/6)/1=cosπ/6=√3/2cosα=邻边/斜边=(s

选项C正确.解析：已知角α终边上一点的坐标是(sin5分之π,cos5分之π)即(cos10分之3π,sin10分之3π）,则由任意角三角函数的定义可知：角α与10分之3π的终边相同所以角α=2Kπ+

（cos5分之π+cos5分之2π+cos5分之3π+cos5分之4π）=（cos5分之π+cos5分之4π++cos5分之2π+cos5分之3π）=2cos[(5分之π+5分之4π)/2]*cos[