求证3sin2a=-4cos2a

因为cot(a/2)-tan(a/2)=cos(a/x2)/sin(a/2)-sin(a/2)/cos(a/2)(通分)=[(cos(a/2))^2-(sin(a/2))^2]/[cos(a/2)si

sin2A+sin2B+sin2C=sin2A+sin2B+sin2(π-A-B)=sin2A+sin2B+sin(2π-2A-2B)=sin2A+sin2B+sin(-2A-2B)=sin2A+si

cos2(a+β)+cos2(a-β)=cos(2a+2β)+cos(2a-2β)=(cos2acos2β-sin2asin2β)+(cos2acos2β+sin2asin2β)=2cos2acos2

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

1）2(sin2α+1)=2*2sinαcosα+2=4sinαcosα+21+sin2a+cos2a=1+2sinαcosα+2cos^2(α)-1=2sinαcosα+2cos^2(α)2(sin

sin2a=2sinacosa1=(sina)^2+(cosa)^22(sin2a+1)=2(sina+cosa)^21+sina2a+cos2a=1+2sinacosa+2(cosa)^2-1=2c

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

由两角和公式展开第一个等式得：根号2（sina+cosa）=sinθ+cosθ,再两边平方,得2(1+sin2a)=1+sin2θ得sin2a=(sin2θ-1)/2;cos2β=1-2sin^2β=

2sin（π/4＋α）=√2(sina+cosa)√2(sina+cosa)=sinθ＋cosθ将这个式子平方,得2(1+sin2a)=1+sin2θ2sin2β=sin2θ2(1+sin2a)=1+

证明：cos^4θ-sin^4θ=(cos²θ-sin²θ)(cos²θ+sin²θ)=(cos²θ-sin²θ)×1=cos²θ

证明：（1）要证sin2A+sin2B+sin2C=2+2cosAcosBcosC成立即证sin2A=2-sin2B-sin2C+2cosAcosBcosC成立又因为2-sin2B-sin2C+2co

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

1/4sin2α4α?

证明:1-cos(2A)=2*[(sinA)^2]1+cos(2A)=2*[(cosA)^2]sin(2A)=2sinA*cosA==>(1+sin2A-cos2A)/(1+sin2A+cos2A)=

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

(1+sin2a)/cos2a=[(cosa)^2+(sina)^2+2sinacosa]/[(cosa)^2-(sina)^2]=(cosa+sina)^2/[(cosa-sina)/(cosa+s

分子=sin²a+cos²a+2siacosa-(cos²a-sin²a)=(sina+cosa)²-(cosa+sina)(cosa-sina)=(

sinθ+cosθ=2sinα(sinθ+cosθ)^2=1+2sinθcosθ=4(sina)^21+2(sinβ)^2=4(sina)^22-cos2β=2-2cos2a2cos2a=cos2β4

sina+cosa=3/4平方得(sina+cosa)²=9/16sin²a+2sinacosa+cos²a=9/161+sin2a=9/16sin2a=-7/16

(sina)^2+(cosa)^2=1所以(sina)^2+4(sina)^2=1(sina)^2=1/53sin2a=6sinacosa=-12(sina)^2=-12/5-4cos2a=-4[1-