cos5π等于多少

多次用倍角公式···可以得到关系关于某个特殊角的多次方···具体自己找找公式···我毕业N久了··公式不记得了···

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

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：是不是如果两个角的度数互余，他们的正弦余

（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[(5分之π-5分之4π)]+2cos[(

只能求正负再问：咋求。。这节课我没听，。。再答：1好像等于37°多吧。忘了。高一学的。然后看图像。再问：哦哦

cosπ/5×cos2π/5＝(2sinπ/5×cosπ/5×cos2π/5)/(2sinπ/5)＝(sin2π/5×cos2π/5)/(2sinπ/5)＝(2sin2π/5×cos2π/5)/2×(2sinπ/5)＝sin4π/5/(4s

不对,左边等于-1/12,右边等于0

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=-√2/4

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

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[cos(π/2)+cosπ/3]=2c

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)-2sin^2(2)+2]=√[cos^2

【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)]=(1/2)sin(π/6)=(1/2)×(1/

cos5π/7=cos(π-2π/7)=-cos2π/7tan5π/7=tan(π-2π/7)=-tan2π/7偶函数,f(-x)=f(x)所以f(cos5π/7)=f(cos2π/7)f(tan5π/7)=f(tan2π/7)第一象限si

(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)(sinπ/12-cosπ/12)=s

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/12)+2*sin(π/12)*cos(π

(sin5π/12-sinπ/12)(cos5π/12+cosπ/12)=(cosπ/12-sinπ/12)(sinπ/12+cosπ/12)=cos²(π/12)-sin²(π/12)=cos(π/6)=(根号3)/2

arccos(cos5π/4)=5π/4arctan(tan4π/3)=4π/3arcsin(sin6)=6

（1）cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)　　=cos(π/5)+cos(2π/5)-cos(2π/5)-cos(π/5)　　=0（2）tan10°+tan170°+sin1866°-sin(-606

π=3.1415926535897932384626433832795

25.12取两位小数