cos^2(π/4-x)-sin^2(π/4-x)在区间[-π/6,π/3]上的最大值与最小值

求证:(sin 2x /(1-cos 2x) )·(sin x /(1+sin x))=tan (π/4-x/2). 化简1+sin x/cos x·sin2x/2cos²（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 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) 化简2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1 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 化简sin（2π+x）cos（π-x）sin（π-x）/cos（x-π）sin（-π-x）sin（π+x） 化简sin^4x-sin^2x+cos^2x 化简:sin^4 x-sin^2 x+cos^2 x sin cos之间怎么转换?sin(π/3-2X）等于cos多少? 化简sin^4x+cos^2x 化简(sin^2x-cos^4x+cos^2x-sin^4x)/(sin^2x-cos^6x+cos^2x-sin^6x) 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 证明 2sin(π+x)cos(π-x)=sin2x