sin2x－cos2x怎么转换为√2sin（2x－兀/4），这个括号里的兀/4不知道怎么来的 sin 2x+cos2x是怎么化简到√2sin(2x+4分之...
sin2x-cos2x-1=\u221a2 sin(2x-\u03c0/4) \u8fd9\u53c8\u662f\u600e\u4e48\u5316\u7684\uff1fsin2x-cos2x-1
=\u221a2(sin2x\u00d7\u221a2/2-cos2x\u00d7\u221a2/2)-1
=\u221a2sin(2x-\u03c0/4)-1
\u5982\u679c\u672c\u9898\u6709\u4ec0\u4e48\u4e0d\u660e\u767d\u53ef\u4ee5\u8ffd\u95ee\uff0c\u5982\u679c\u6ee1\u610f\u8bb0\u5f97\u91c7\u7eb3
\u5982\u679c\u6709\u5176\u4ed6\u95ee\u9898\u8bf7\u91c7\u7eb3\u672c\u9898\u540e\u53e6\u53d1\u70b9\u51fb\u5411\u6211\u6c42\u52a9\uff0c\u7b54\u9898\u4e0d\u6613\uff0c\u8bf7\u8c05\u89e3\uff0c\u8c22\u8c22\u3002
\u795d\u5b66\u4e60\u8fdb\u6b65
\u8f85\u52a9\u89d2\u516c\u5f0f
用辅助角公式：asina+bcosa=√(a²＋b²)sin(a+φ)sin2x－cos2x
=√2×[(√2/2)sin2x－(√2/2)cos2x]
=√2[sin2xcos(π/4)－cos2xsin(π/4)]
=√2sin(2x－π/4)
sin2x－cos2x
=√2×[(√2/2)sin2x－(√2/2)cos2x]
=√2[sin2xcos(π/4)－cos2xsin(π/4)]
=√2sin(2x－π/4)
aSINx bCOSx=sqrt(a^2 b^2)SIN(x 角度)。角度=artanb/a
绛旓細sin2X-cos2X =鈭2sin2Xcos蟺/4-鈭2cos2Xsin蟺/4 =鈭2sin锛2x-蟺/4锛夊嶈鍗婅鍏紡锛歴in ( 2伪 ) = 2sin伪 路 cos伪 sin ( 3伪 ) = 3sin伪 - 4sin & sup3 ; ( 伪 ) = 4sin伪 路 sin ( 60 + 伪 ) sin ( 60 - 伪 )sin ( 伪 / 2 ) = 卤 鈭( ( 1 - cos伪...
绛旓細瑙ｅ寲绠鏄牴2sin(x-鍏/4)
绛旓細sin2x鈥攃os2x =鈭2[sin2xcos(蟺/4)鈥攃os2xsin(蟺/4)]=鈭2sin(2x-蟺/4)
绛旓細=sin2x-cos2x =鈭2.sin(2x-蟺/4)
绛旓細鐢ㄨ緟鍔╄鍏紡锛歛sina+bcosa=鈭(a²锛媌²)sin(a+蠁)sin2x锛峜os2x =鈭2脳[(鈭2/2)sin2x锛(鈭2/2)cos2x]=鈭2[sin2xcos(蟺/4)锛峜os2xsin(蟺/4)]=鈭2sin(2x锛嵪/4)
绛旓細f(x)=sin2x-cos2x =鈭2*[sin2x*(鈭2)/2 - cos2x*(鈭2)/2]=鈭2*[sin2x*cos(蟺/4) -cos2x*sin(蟺/4)]=鈭2*sin(2x-蟺/4)
绛旓細sin2X=2sinXcosX cos2X=cosX^2-sinX^2 cosX^2+sinX^2=1 浠ｅ叆鍏紡鏈 鍘熷紡=2sinXcosX-锛坈osX^2-sinX^2锛=cosX^2+sinX^2=1 鎵浠ユ湁2sinXcosX=2cosX^2 鎵浠ユ湁cosX鎴栦笉绛変簬0涓ょ鎯呭喌 褰揷osX=0鏃讹紝 sinX=1鎴-1 姝ゆ椂X绛変簬 2k蟺+蟺/2 鎴(2k+1)蟺+蟺/2 褰揷osX涓嶇瓑浜0鏃 co...
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