cosΨ cos2Ψ cos3Ψ - cosnΨ＝

sinacosacos2a=1/2sin2acos2a=1/4sin4a再问：最后一步怎么得出来的啊？？再答：=1/4(2sin2acos2a)=1/4sin(2x2a)=1/4sin4a

[sin4α/(1+cos4α)]×[cos2α/(1+cos2α)]×[cosα/(1+cosα)]=[2*2*[2sin(α/2)cos(α/2)]cosα/(2cos2α)]×cos2α/[2(

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

cosx+cos2x+.cos(nx)=sin(nx/2)cos((n+1)x/2)/sin(x/2)sinx+sin2x+.sin(nx)=sin(nx/2)sin((n+1)x/2)/sin(x/

(sinα+sin2α+sin3α)/(cosα+cos2α+cos3α)=(2sin2αcosα+sin2α)/(2cos2αcosα+cos2α).此处使用和差化积即可=(1+2cosα)sin2

sin@cos@cos2@＝sin2@cos2@/2=sin4@/4

cos3a=cos(2a+a)=cos2acosa-sin2asina=(2cos²a-1)cosa-2sin²acosa=2cos2acosa-cosa∴1+cosa+cos2a

利用黄金分割容易得COSπ/5=(根号5+1)/4COS2π/5=(根号5-1)/4乘起来是1/4

原式=[(2cos²α-1)-(2cos²β-1)]/(cosα-cosβ)=2(cos²α-cos²β)/(cosα-cosβ)=2(cosα-cosβ)(c

cosπ/5×cos2π/5=(2sinπ/5cosπ/5×cos2π/5)/2sinπ/5=(2sin2π/5×cos2π/5)/(4sinπ/5)=(sin4π/5)/(4sinπ/5)=1/4

证明cosα+cos3α=2*cos((α-3*α)/2)cos((α+3*α)/2)=2cosα*cos(2α)1+cos(2α)=2(cosα)^2上面两个式子相加有1+cosα+cos2α+co

cosθ=[e^(iθ)+e^(-iθ)]/2cos2θ=[e^(2iθ)+e^(-2iθ)]/2cos3θ=[e^(3iθ)+e^(-3iθ)]/2...cosnθ=[e^(inθ)+e^(-inθ

COS1+COS2+COS3=0移项平方得（COS1+COS2）^2=(-COS3)^2得：(COS1)^2+(COS2)^2+2*COS1*COS2=(COS3)^2SIN1+SIN2+SIN3=0

正切,余弦全化为正弦

cos1°²+cos2°²+cos3°²+cos4°²...+cos89°²＝cos1°²+cos2°²+cos3°²+

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

原式=2sinπ/5cosπ/5cosπ/5/(2sinπ/5)=sin2π/5cosπ/5/(2sinπ/5)=2sin2π/5cosπ/5/(4sinπ/5)=sin4π/5/(4sinπ/5)=

cos²(α+β)=0.5[cos(2α+2β)+1]=0.5os2αcos2β-0.5sin2αsin2β+0.5cos²(α-β)=0.5[cos(2α-2β)+1]=0.5c

用acos3a=cos(2a+a)=cos2acosa-sin2asina=(2cos²a-1)cosa-2sinacosa*sina=2cos³a-cosa-2(1-cos&su

cos3α＝4(cosα)^3－3cosαcos2α＝2(cosα)^2－1cosαcos2αcos3α=cosα*[4(cosα)^3－3cosα]*[2(cosα)^2－1]=8(cosα)^6-