√1-sin ^2(3pai /5)化简的结果 我想知道答案为什么是-√cos (3pai /

已知cos(a-pai/6)+sina=4√3/5,则sin(a+7/6pai)的值是? [tan(pai-a)cos(2pai-a)sin(-a+3pai/2)]/[cos(-a-pai)sin(-pai-a 已知sin（pai－a）－cos（pai＋a）＝√2/3 （pai/2 sin（x+pai/3)+2sin(x-pai/3)-3^1/2 cos(2pai/3-x)化简得〇, cos(pai/15)×cos(2pai/15)×cos(3pai/15)×cos(4pai/15)×cos(5pai/ 试求y=1/2cos(paix+pai/3)-sin(paix+5pai/6)的单调增区间 化简!f(x)=sin(pai-x)cos(3/2pai+x)+sin(pai+x)sin(3/2pai-x) 化简：[cos(a+pai)*sin^2(a+pai)]/[tan^2(pai+a)*cos^3a] f(a)=[sin(pai-a)cos(2pai-a)tan(-a+3pai/2)]/[cos(-pai-a)]则f(- 已知cos (a+pai /3)=sin (a-pai /3),则tan a的值是 cosx -（根号3）*sinx这道题的答案是-2sin(x +pai /3)还是2sin(x +pai /3) sin(pai/6+a)=1/3,则cos(2pai/3-2a)=?