sina(丌/6+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)化简并计算

sina(丌/6+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)化简并计算 化简并计算sina(丌/6+2k丌).cos(11/3丌)-tan(5/4丌+2k丌) sina(a/b+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)化简并计算 sina(a/b+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)化简并计算,具体怎么解? sin(a+kπ)=2cos[a+(k+1)π].求5cosa+3sina/4sina-2cosa sin(a+kπ)=2cos[a+(k+1)π].求5cosa+3sina/4sina-2cosa 当sina+sina*2=1时,COS*2+COS*6的值 已知sin(a+Kπ)=-2cos（a+Kπ）,K∈Z则 4sina-2coxa/5cosa+3sina sin(a+kpai)=-2cos(a+kpai),k属于Z求(4sina-2cosa)/(5cosa+3sina) 已知sin(kπ+a)=-3cos(kπ+a)(k∈z),则(4sina+cosa)/(2sina-cosa)= 已知sin(a+kpai)=-2cos(a+kpai)(k属于z)求(4sina-2sina)/(5sina=3sina)1.(4sina-2sina)/(5sina=3sina)2.0.25sin^2 a+0.4cos^2 a在线等,需要过程,有答案了会加分求(4sina-2cosa)/(5cosa+3sina) 之前打错 若sina=1/3,求cos(（2k+1/2）π+a)+cos(（2k-1/2）π-a)cos[（2k+1）π/2+a]+cos[（2k-1）π/2-a] Sina+sin^2a=1,cos^2a+cos^s4a+cos^6a sina+sina^2=1,化简cosa^2+cos^6+cosa^8 -sina/-cos*-sina*cosa 求高手``若sin(a+k∏)=-2cos(a+k∏)(k∈Z),求1/4sina+2/5(cosa)^2 若sin(a+k∏)=-2cos(a+k∏)(k∈Z),求1/4sina+2/5(cosa)^2 证明 cosa/(1+sina0-sina/(1+cosa)=2(cosa-sina)/(1+sina+cos)cosa/(1+sina）-sina/(1+cosa)=2(cosa-sina)/(1+sina+cos)