cosΠ 7 cos3Π 7 cos5Π 7

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

(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

sin25π/6+cos5π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(-5π/6)×cos(-5π/3)=sin(4π+π/6)+cos(2π-π/3)+ta

cosπ/7-cos2π/7+cos3π/7=cosπ/7+cos3π/7+cos5π/7=(2sinπ/7*cosπ/7+2sinπ/7*cos3π/7+2sinπ/7*cos5π/7)/(2sin

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/

cos²5π/12+cos²π/12+cosπ/12*cos5π/12由于cos5π/12=sin（π/2-5π/12）=sinπ/12原式=sin²π/12+cos&#

cos^2(π/4+π/6)+cos^2(π/4-π/6)+1/2[cos(5π/12+π/12)+cos(5π/12-π/12)]=cos^2(π/4+π/6)+cos^2(π/4-π/6)+1/2

解本题就是证明cos^2（75°）+cos^2（15°）+cos（75°）cos（15°）=5/4我们用分析法证明欲证cos^2（75°）+cos^2（15°）+cos（75°）cos（15°）=5/

你画图就行了啊..把sin,cos,tan的图画出来就看出来了..再问：画不来才来问的……再答：额。。。再答：先做一个直角坐标系再答：sin(0)=0sin(90)=1sin(180)=0sin(27

①-√2/2②1/2③同①④1⑤√3/3⑥同①

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,

cos（6π-α）=√3/3-->cosα=√3/3然后就降幂化简原式,但你的题目好像没写全

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

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

（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[

第一个信号的周期是1,第二个信号的周期是2/3第三个信号的周期是2/5那么这个三个信号相加的周期应是这三个信号周期的最小公倍数,即2角频率与周期的关系是：w=2πf=2π/T=π再问：加起来不是整数啊