sin^2(B C) cos^2B cos^2C sinBsinC大于等于2

已知向量a=(cosα,sinα)b=(cosβ,sinβ)|a+b|=2|a-b|若0＜α＜π/2,-π/2＜β＜0,且sinβ=-5/13求sinα解析：∵向量a=(cosα,sinα)b=(co

[sin(2A+B)]/sinA-2cos(A+B)=[sin(A+B+A)]/sinA-2cos(A+B)=[sin(A+B)cosA+cos(A+B)sinA]/sinA-2cos(A+B)=[s

证明：输入过于麻烦,用换元法吧设A=sin²A,B=sin²B∵sin^4a/sin^2b+cos^4a/cos^2b=1即A²/B+(1-A)²/(1-B)=

如题,原式=sin平方a+sin平方b-sin平方a*sin平方b+cos平方a*cos平方b=sin平方a+sin平方b-sin平方a*(1-cos平方b)+cos平方a*cos平方b=sin平方a

最后=1具体过程我打给你了这里打不下

Sin^2A+Sin^2B-Sin^2ASin^2B+Cos^2ACos^2B=Sin^2A(1-Sin^2B)+Sin^2B+Cos^2ACos^2B=Sin^2ACos^2B+Cos^2ACos^

(2sina+cosa)=-5(sina-3cosa)7sina=14cosasina=2cosa

a+b=a+b/2+b/2,a=a+b/2-b/2sin(a+b)-sina=sin[（a+b/2）+(b/2)]-sin[（a+b/2）-(b/2)]展开=sin（a+b/2）*cosb/2+cos

利用公式把变量的形式化简得越少越好(sina)^2*(sinb)^2+(cosa)^2*(cosb)^2-1/2cos2a*cos2b=(1-(cosa)^2)(1-(cosb)^2)+(cosa)^

向量AC=（cosα-2,sinα）,向量BC=（cosα,sinα-2）因向量AC垂直于BC所以向量AC*.向量BC=0即（cosα-2,sinα）*.（cosα,sinα-2）=0cos^2α-2

1.cos（x+B）*cos（x-B）=(cosxcosB-sinxsinB)(cosxcosB+sinxsinB)=(cosx)^2(cosB)^2-(sinx)^2(sinB)^2=(cosx)^

sin^2A-cos^2B=0sin^2A=cos^2B=[sin(90-B)]^2所以,A=90-B,或A=-(90-B)即：A+B=90,或A-B=90所以,cos(A+B)=0

cos^2a-sin^2b=(1+cos2a)/2-(1-cos2b)/2=(cos2a+cos2b)/2=cos(a+b)cos(a-b)=1/3

1+sin2B=(sinB+cosB)^2cos^2B-sin^2B=(cosB+sinB)(cosB-sinB)所以1+sin2B/cos^2B-sin^2B=sinB+cosB/cosB-sinB

sin^2/(sin-cos)-(sin+cos)/(tan^2-1)=sin^2/(sin-cos)-(sin+cos)/[(sin^2/cos^2)-1]=sin^2/(sin-cos)-(sin

原题是这样子吧：cos(a+b)cos(a-b)=1/5,则(cosa)^2-(sinb)^2=?cos(a+b)cos(a-b)=(cosacosb-sinasinb)(cosacosb+sinas

sinx+cosx=(√2)*(√2/2)(sinx+cosx)=√2(sinxcos45°+cosxsin45°)=√2sin(x+45°)公式:asinx+bcosx=【√(a²+b&#

sin(A+B)-sinA=2cos(A+B/2)sin(B/2)和差化积sin(A+B)-sinA=1/2[cos(A+B)/2]*[sin(A+B-A)/2]=2cos(A+B/2)sin(B/2

A+B+C=180A+B=180-C(A+B)/2=90-C/2sin(A+B)/2=sin[90-C/2]=cosC/2

证明：在△ABC中∵A+B+C=180°∴A/2+(B+C)/2=90°∴sinA/2=sin[90-(B+C/2}=cos(B+C)/2