cosα cos3α=2cos2αcosα

[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/

cos2β=1-2sin²β=1-2sinθcosθ=1-sin2θ因为4sin²α=1+2sinθcosθ=1+sin2θ得sin2θ=4sin²α-1所以cos2β=

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

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

用a和b左边=cos[(a+b)+(a-b)]cos[(a+b)-(a-b)]=[cos(a+b)cos(a-b)-sin(a+b)sin(a-b)][cos(a+b)cos(a-b)+sin(a+b

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

(2sin2α/1+cos2α)*(cosα)^2/cos2α=2tana*(cosα)^2/cos2α=2tana*(cosα)^2/cos2α=2sinα*cosα/cos2α=sin2α/cos

sinα-cosα=根号2/2平方得到：1-2sinacosa=1/2sinacosa=1/4sina+cosa=±√6/2sin3次方α+cos3次方α=(sina+cosa)(1-sinacosa

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

由tan(α+β)/2=√2/2,有tan(α+β)=2√2cos（2α+2β）=[cos²(α+β)-sin²(α+β)]/[cos²(α+β)+sin²(α

原式子=[cos2(a+b)+cos2(a-b)]/2-[1+cos2(a-b)]/2=[cos2(a+b)-1]/2=-[sin(a+b)]^2sin(a+b)=2tan(a+b)/2/(1+tan

∵2sin²α-cos²α+sinαcosα-6sinα+3cosα=0==>(sinα+cosα)(2sinα-cosα)-3(2sinα-cosα)=0==>(2sinα-co

∵sinα+cosα=根号2∴1+2sinα*cosα=2sinα*cosα=1/2sin3α+cos3α=(sinα+cosα)(sin^2α-cosαsinα+cos^2α)=√2*(1-sinα

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

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

2sinα+cosα=-5sinα+15cosα7sinα=14cosαtanα=0.5cos2α=((cosα)^2-(sinα)^2)/((cosα)^2+(sinα)^2)=(1-(tanα)^