设cos(a-p/2)=-1/9,sin(a/2-p)=2/3,且派/2＜a＜派,0＜p＜派/2,则cos(a+p)=

设cos(a-p/2)=-1/9,sin(a/2-p)=2/3,且派/2＜a＜派,0＜p＜派/2,则cos(a+p)= 由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证：1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a) 设tan a=-1/2,计算1/(sin^2a-sin a cos a-2cos^2a) 设函数f(x)=p·q,其中向量p=(sin x,cos x+sin x),q=(2cos x,cos sin x) ( 求证1+sin a+cos a+2sin acos a/1+sin a+cos a=sin a+cos a 设sinα+sinβ=1/3,则sinα-cos方β的最大值是A 4/3 B 4/9 C -11/12 D -2/3cos方β=cos的平方β 求证:sin 2a/1+sin a+cos a=sin a+cos a-1急 设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π＋α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2（-sinα)(-cosα）+cosα/[1+sin² 设cos(a-1/2b)=-1/9,sin(1/2a-b)=2/3,且π/2 设cos(a-β/2)=-1/9,sin(a/2-β)=2/3,且π/2 设锐角三角形ABC中,2sin^2A-cos^2A=2.(1)求 设cos2a=1/2 则cos^4a+sin^4a= 设tan=2 求1+2sinacosa/sin平方a-cos平方a急 设sin(π/4-a/2)cos(π/4-a/2)=1/6,求sin2a 请问数学题,设sin a+cos a=3/5,则sin 2a=? 设sinα-cosα=1/2,求sinα*cosα= 设f(a)=2sin(π+a)cos(π-a)-cos(π+a)/1+sin^2a+cos(3π/2+a)-sin^2(π/2+a),且1+2sina≠0,化简f(a) 设角a=-35π/6,2sin(兀+a)cos(兀-a)-cos(兀+a)/1+sin^2a+sin(兀-a)-cos^2(兀+a)值等于