cos2πx等于多少度

∵f（x）=cos2(x−π4)−cos2(x+π4)=1+cos(2x−π2)2-1+cos(2x+π2)2=sin2x2-−sin2x2=sin2x．∴T=2π2=π，又f（-x）=sin（-2x

解由cos2α+sin^2α=1-2sin²α+sin²α=1-sin²α=cos²a由sin²α+cos²a=1,两边除以cos²

(sinа+cosа）²/cos2а=(sin²a+2sina·cosa+cos²a)/(cos²-sin²a)=(tan²a+2tana+

原式=sinπ7(cos2π7+cos4π7+cos6π7)sinπ7=sinπ7cos2π7+sinπ7cos4π7+sinπ7cos6π7sinπ7=12(sin3π7−sinπ7)+12(sin

f(sin(x/2))=cosx+1=1-2(sin(x/2))^2+1=2-2(sin(x/2))^2令y=sin(x/2)则f(y)=2-2y^2令y=cos(x/2)f(cos(x/2))=2-

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

负得25分之7哦.在那个帖里,已经写过程了哦.

先把它看成分母为1的分数,（cosπ/5）（cos2π/5)/1,然后分子分母同时乘sin（π/5）,这样分子上可以用一下sin的二倍角公式式子变为：sin（2π/5）*cos（2π/5）/2sin(

诱导公式:sin（π/2+Θ）=cosΘ=5分之3二倍角公式：cos2Θ=2（cosΘ的平方）-1=-25分之7

这一类题可用对称法：设sinπ/5×sin2π/5=mcosπ/5×cos2π/5＝n则4mn=(2sinπ/5×cosπ/5)×(2sin2π/5×cos2π/5)=sin2π/5×sin4π/5=

cos^2(π/2-a)+cos2^2(π/6+a)=sin^2a+1/2(1+cos(π/3+2a)=1/2(1-cos(π/3+2a)+1/2(1+cos(π/3+2a)=1再问：cos^2(3/

(1)a·b=(cos2/3x,sin2/3x)*(cos2/x,-sin2/x)=cos2/3x*cos2/x-sin2/3x*sin2/x=cos(2/3x+2/x)=cos8/3x|a+b|=√

1f(x)=a·b+2λ|a+b|a·b=(cos(3x/2),sin(3x/2))·(cos(x/2),-sin(x/2))=cos(2x)|a+b|^2=|a|^2+|b|^2+2a·b=2+2c

a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2),(1)a*b=(cos3x/2,sin3x/2)*(cosx/2,-sinx/2)=cos(3x/2)*cos(x/2)-

（Ⅰ）f(x)＝sin2ωx−cos2ωx−1＝2sin(2ωx−π4)−1．因为T2＝π2，所以T=π，ω=1．（3分）所以f(x)＝2sin(2x−π4)−1．所以f(π4)＝0（7分）（Ⅱ）f(

令：a+π/3=bcos2b=1-2(sinb)^2=1-(1/3)^2=7/9

解cos2θ=1-2sin^2θ=1-2×(5分之3)平方=1-2×25分之9=1-25分之18=25分之7

诱导公式:sin（2分之π+Θ）=cosΘ=5分之3二倍角公式：cos2Θ=2（cosΘ的平方）-1=-25分之7

(Sin2※)-(tan※)*(cos2※)等于多少呢?(假设※　是阿尔法吧.)知道手机网友你好：你要发布问题,就把问题发完整.问的题目是什么,写清楚.以免浪费短信费,耽误你.

tan(π/4－a)=[1－tana]/[1＋tana]=3,则tana=－1/2.而sin2a－coa2a=[2sinacosa－cos²a＋sin²a]/[sin²a