以知cos(a+β )=-1,求证sin(2a+β )+sinβ=0

以知cos(a+β )=-1,求证sin(2a+β )+sinβ=0 求证:sin a (1+cos2a)=sin a cos a 求证1+sin a+cos a+2sin acos a/1+sin a+cos a=sin a+cos a 求证:sin 2a/1+sin a+cos a=sin a+cos a-1急 求证：tan a/2＝(1-cos)/sin a=sin a/(cos a+1) 求证 (cos a÷(1-sin a))=((1+sin a)÷cos a) 求证 1+sin a+cos a+2sin a*cos a/1 +sin a +cos a = sin a+ cos a 有什么思路! 求证：sin²a+sin²β-sin²a×sin²β+cos²a×cos²β=1x是乘以 已知sinα=a*sin(α+β) (a>1),求证:tan(α+β)=sinβ/(cosβ-a) 求证sin^4a+cos^4a=1 求证:sin^2/(sin-cos) - (sin+cos)/(tan^2 -1) =sin+cos 求证:sin a(1+tan a)+cos a(1+1/tan a)=1/sin a+1/cos a 求证sin^2α+sin^2β-sin^2αsin^2β+cos^2cos^2β=1 高一向量问题.已知向量a=(cosα,sinα),向量b=（cosβ,sinβ）,向量c=（cosγ,sinγ）已知向量a=(cosα,sinα),向量b=（cosβ,sinβ）,向量c=（cosγ,sinγ）且3cosα+4cosβ+5cosγ=0, 3sinα+4sinβ+5sinγ=0.（1）求证向量a 已知cos^4β/cos^2a+sin^4β/sin^2a=1,求证cos^4a/cos^2β+sin^4a/sin2^β=1跪谢,我一定会追加50分的! 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 求证2cos^2a+sin^4a=cos^4a+1 求证3-sin^4a-cos^4a/2cos^a=1+tan^a+sin^a 速答 谢谢