sin4a-cos4a=1-2cos2a

1.倍角公式：cos2α=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)3+cos4a-4cos2a=3+（2cos^2(2a)-1)-4(1-2sin^2(a))

左边=sin^2(2a)+cos^2(2a)-2sin2a*cos2a=1-2sin2a*cos2a=右边.用到的公式有sin^2(A)+cos^2(A)=1和sin2A=2sinAcosA

sin(π/4+a)=sin(π/4)*cosa+sina*cos(π/4)sin(π/4-a)=sin(π/4)*cosa-sina*cos(π/4)原式=[sin(π/4)*cosa+sina*c

1、tan2a=[tan(a+b)+tan(a-b)]/[1-tan(a+b)·tan(a-b)]=(3+2)/(1-3*2)=-1∴sin4a=2tan2a/[1+(tan2a)^2]=2(-1)/

上面=sin2a-sin4a+cos2a-cos4a=sin2a(1-sin2a)+cos2a(1-cos2a)=2sin2acos2a下面=sin2a-sin6a+cos2a-cos6a=sin2a

解sin(π/2+a-π/4)sin(π/4-a)=cos(a-π/4)sin(π/4-a)=cos(π/4-a)sin(π/4-a)=1/6∴sin[2(π/4-a)]=1/3即sin(π/2-2a

（sin2a+cos2a-1)(sin2a-cos2a+1)=sin²2a-（cos2a-1）²=-2cos²2a+2cos2a=根号3sin4a,即-2cos²

左边平方差=(sin²a+cos²a)(sin²a-cos²a)因为sin²a+cos²a=1所以左边=sin²a-cos&sup

[cos(4a-π/2)*sin(π/2-2a)]/[(1+cos2a)*(1+cos4a)]=[cos(π/2-4a)*sin(π/2-2a)]/[(1+cos2a)*(1+cos4a)]=(sin

sina+cosa＝根号2a=45度sin4a-cos4a=sin180-cos180=0-（-1）=1

cos2a=根号2/3sin2a=±根号5/3sin4a+cos4a=2cos²2a-1+2sin2acos2a=-1/9±根号10/9

原式=sin4a*cosa*cos2a/[(1+cos2a)(1+cos4a)*cosa〕=sin4a*cos2a/〔2cosacosa*2cos2acos2a〕=2sin2acos2a/（4cosa

sina+sin2a+sin4a+sin5a=2sin3acos2a+2sin3acosa=2sin3a(cos2a+cosa)cosa+cos2a+cos4a+cos5a=cosa+cos5a+co

这个等式是不成立的.假设a=30,那么sin4a-cos4a=sin120-cos120=sin60+cos60=/2sin2a-cos2a=sin60-cos60=/2

4cosAcosB=√6,4sinAsinB=√2--->cosAcosB=√6/4,sinAsinB=√2/4.(1-cos4A)(1-cos4B)={1-[1-2(sin2A)^2]}*{1-[1

cos4a=2(cos2a)²-1=2(1-2(sina)²)-1cos4b=2(cos2b)²-1=2(1-2(sinb)²)-12cos4a-1/2cos4

证：∵（sin2a-cos2a)^2=sin²2a+cos²2a-2sin2acos2a而sin²2a+cos²2a=1,2sin2acos2a=sin4a,∴

根据你提供的第一个等式,再由sina^2+cosa^2=1,可以知道sina=cosa^2,同时根据等式可以求出sina=（√5-1）/2,另一值舍去了.因为sina=cosa^2,所以sina^2=

(sin2a-cos2a)^2=sin2a^2+cos2a^2-2*sin2a*cos2a=1-2*sin2a*cos2a=1-sin4a因为sin2a^2+cos2a^2=12*sin2a*cos2

答：sina+cosa=√2两边平方得：sin²a+2sinacosa+cos²a=22sinacosa=1sin2a=1,cos2a=0sin4a+cos4a=2sin2acos