sin(Π/2 - x)=cos(Π/4 - x/2)、以及 1-cos(Π/2 - x)=sin(Π/4 - x/2)

求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2). 已知sin[a-b]cos a-cos[b-a]sin a=3/5,b是第三象限角,求sin[b+5π/4]的值第一题1/2cos x-√3/2sin x第二题√3sin x+cos x第三题√2[sin x-cos x]第四题√2cos x-√6sin x cos (2x-π/3)=2sinπ/4 化简sin(x+7π/4)+cos(x-3π/4）步骤我已经找到撒sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2- Sqrt[2] * { Sin( x + π/4 ) * cos[ π/4 ] + Cos[ x + π/4] * sin[ π/4 ] = Sqrt[2] *sin (x + π/2)我这一步看不懂.. 求证:sin(π/4+x)/sin(π/4-x)+cos(π/4+x)/(π/4-x)=2/cos2xsin(π/4+x)/sin(π/4-x)+cos(π/4+x)/cos(π/4-x)=2/cos2x 证明 2sin(π+x)cos(π-x)=sin2x f(x)=2sin^2(π/4+x)+根号3(sin^2x+cos^2x)f(x)=2sin^2(π/4+x)+根号3(sin^2x-cos^2x) 符号打错了 若cosx=2/3,x是第四象限角,求sin(x-2π)+sin(-x-3π)cos(x-3π)/cos(π-x)-cos(-π-x)cos(x-4π)的值 cos(2x-π/2)=sin 2x sin(x+1/2π)=cosx?还是cos-x 函数y=sin(x)cos(x+ π/4)+cos(x)sin(x+π/4)的最小正周期是 f(x)=sin(x+π/4)图像怎样变成F(x)=cos(2x) 求函数y=cos(9/2π+x)+sin^2x的最大值和最小值还有一题 已知sinx+cosx=1/2,求sin^3x+cos^3x和sin^4x+cos^2x f(x)=cos(2x-x/3)+2sin(x-π/4)sin(x+π/4)sin(x+π/4)化简. sin(Π/2 - x)=cos(Π/4 - x/2)、以及 1-cos(Π/2 - x)=sin(Π/4 - x/2) y=2cos(x/4+ π/3)-3sin(x- π/4)y=sin(4x- π/3)-cos(3x+π/2) 已知f(x)=sin(π/6-x)^2-cos(π/4+x)^2+cos(π/6)cos(π/6-2x)化简f(x)