求证:sin(a+β)sin(a-β）=（cosβ）^2-(cosa)^2

求证:sin(a+β)sin(a-β）=（cosβ）^2-(cosa)^2 已知sinα=a*sin(α+β) (a>1),求证:tan(α+β)=sinβ/(cosβ-a) 以知cos(a+β )=-1,求证sin(2a+β )+sinβ=0 求证sin（360-a）=-sina 已知3sinβ=sin(2a+β）,求证：tan(a+β）=2tana 求证：sin²a+sin²β-sin²a×sin²β+cos²a×cos²β=1x是乘以 求证恒等式sin(A+B)sin(A-B)=sin²A-sin²B 求证:sin(a+b)sin(a-b)=sin^2a-sin^2b 求证:sin a (1+cos2a)=sin a cos a 3 在三角形ABC中,已知（a2+b2）sin（A-B）=（a2-b2）sin(A+B) 求证：ABC是等腰或直角三角形(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B) sin^A*(sin(A+B)-sin(A-B))=sin^B*(sin(A-B)+sin(A+B)) sin^A*2c 在△ABC中,求证；sin^(A/2)+sin^(B/2)+sin^(C/2)=1-2sin(A/2)sin(B/2)sin(C/2） 已知：α、β是锐角,a*sinα+b*cosβ=sinβ,a*sinβ+b*cosα=sinα,tan（（α+β）/2）=a+1,求证：a^2+b=1 sin(a+b)sin(a-b)=sin2a-sin2b求证 求证：sin(A+B)sin(A-B)=sinAsinA-sinBsinB 求证:(tan a*sin a)/(tan a-sin a)=(tan a+sin a)/(tan a*sin a) 若2sin(π/4+a）=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0. 若2sin(π/4+a）=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0 求证cos(a+β)cos(a-β)=(cosa)^2-(sinβ)^2 高一数学1）求证cos(a+β)cos(a-β)=(cosa)^2-(sinβ)^22)求证 sin(a+β)sin(a-β)=(sina)^2-(sinβ)^2