cos5/4π等于多少

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

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：

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

（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°+tan17

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

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

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

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

【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/

cosπ/5×cos2π/5＝(2sinπ/5×cosπ/5×cos2π/5)/(2sinπ/5)＝(sin2π/5×cos2π/5)/(2sinπ/5)＝(2sin2π/5×cos2π/5)/2×(

你对着四个象限就行了,4在第三象限,sin4小于0,5在第四象限,cos5大于0,8-6.28＝1.72＞3.14/2,所以,8在第二象限,tan8小于0,(-3)在第三象限,tan(-3)大于0.

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

①-√2/2②1/2③同①④1⑤√3/3⑥同①

点P横坐标为正,纵坐标为负,在第四象限!P（1/2,-根3/2）α终边与-60度终边相同.

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[

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