sin(2x)-cos(2x)=0 这题怎么做? sinxcos2x等于？
sin2x-cos2x=0\u56fe\u600e\u4e48\u753bsin2x-cos2x=0⇒sin2x=cos2x⇒tan2x=1\uff08x\u2260\u03c0/2\uff09\u4e3a\u4e00\u4e2a\u65b9\u7a0b\uff0c\u82e5\u771f\u7684\u6709\u56fe\u50cf\u7684\u8bdd\uff0c\u4e5f\u8be5\u4e3a\u4e00\u4e2a\u70b9\u3002\uff08\u4f60\u662f\u4e0d\u662f\u8981\u95eesin2x-cos2x=f\uff08x\uff09\u7684\u56fe\u50cf\u600e\u4e48\u753b\u554a\uff1f\uff09
\u89e3\u9898\u8fc7\u7a0b\u5982\u4e0b\uff1a
sinxcos2x
=2cosx\u00d7(cosX)^2-(sinX)^2
=2cosx\u00d72\u00d7(cosX)^2-1
=2cosx\u00d71-2\u00d7(sinX)^2
\uff1d1/2(sinxcos2x\uff0bcosxsin2x\uff0bsinxcos2x\uff0dcosxsin2x)
\uff1d1/2(sin(x\uff0b2x)\uff0bsin(x\uff0d2x)
\uff1d1/2(sin3x\uff0dsinx)
\u6269\u5c55\u8d44\u6599\u516c\u5f0f\uff1a
\u53ea\u4f7f\u7528\u51e0\u4f55\u548c\u6781\u9650\u7684\u6027\u8d28\uff0c\u53ef\u4ee5\u8bc1\u660e\u6b63\u5f26\u7684\u5bfc\u6570\u662f\u4f59\u5f26\uff0c\u4f59\u5f26\u7684\u5bfc\u6570\u662f\u8d1f\u7684\u6b63\u5f26\u3002\uff08\u5728\u5fae\u79ef\u5206\u4e2d\uff0c\u6240\u6709\u89d2\u5ea6\u90fd\u4ee5\u5f27\u5ea6\u6765\u5ea6\u91cf\uff09\u3002\u6211\u4eec\u53ef\u4ee5\u63a5\u7740\u4f7f\u7528\u6cf0\u52d2\u7ea7\u6570\u7684\u7406\u8bba\u6765\u8bc1\u660e\u4e0b\u5217\u6052\u7b49\u5f0f\u5bf9\u4e8e\u6240\u6709\u5b9e\u6570x\u90fd\u6210\u7acb\uff1a
\u8fd9\u4e9b\u6052\u7b49\u5f0f\u7ecf\u5e38\u88ab\u7528\u505a\u6b63\u5f26\u548c\u4f59\u5f26\u51fd\u6570\u7684\u5b9a\u4e49\u3002\u5b83\u4eec\u7ecf\u5e38\u88ab\u7528\u505a\u4e09\u89d2\u51fd\u6570\u7684\u4e25\u683c\u5904\u7406\u548c\u5e94\u7528\u7684\u8d77\u70b9\uff08\u6bd4\u5982\uff0c\u5728\u5085\u91cc\u53f6\u7ea7\u6570\u4e2d\uff09\uff0c\u56e0\u4e3a\u65e0\u7a77\u7ea7\u6570\u7684\u7406\u8bba\u53ef\u4ece\u5b9e\u6570\u7cfb\u7684\u57fa\u7840\u4e0a\u53d1\u5c55\u800c\u6765\uff0c\u4e0d\u9700\u8981\u4efb\u4f55\u51e0\u4f55\u65b9\u9762\u7684\u8003\u8651\u3002\u8fd9\u6837\uff0c\u8fd9\u4e9b\u51fd\u6570\u7684\u53ef\u5fae\u6027\u548c\u8fde\u7eed\u6027\u4fbf\u53ef\u4ee5\u5355\u72ec\u4ece\u7ea7\u6570\u5b9a\u4e49\u6765\u786e\u7acb\u3002
看图解题过程：
sin2x=cos2x
tan2x=1
2x=kπ十π/4
x=kπ/2十π/8
绛旓細f(x)=sin2x鈥cos2x=sin2xcos(蟺/4)鈥攃os2xsin(蟺/4)=鈭2sin(2x-蟺/4)鍥犱负鍑芥暟y=sinx鐨勫懆鏈熸槸蟺 鎵浠(x)鍗曡皟閫掑鍖洪棿锛(2k蟺-蟺/2)鈮(2x-蟺/4)<(2k蟺+蟺/2) (k涓烘暣鏁)鍗(k蟺-蟺/8)鈮<(k蟺+3蟺/8)f(x)鍗曡皟閫掑噺鍖洪棿锛(2k蟺+蟺/2)鈮(2x-蟺/4)<(2k蟺+3蟺/2) ...
绛旓細y=sin2x-cos2x y'=(sin2x-cos2x)'=(sin2x)'-(cos2x)'=cos2x*(2x)'+sin2x(2x)'=2cos2x+2sin2x =2鈭2(鈭2/2cos2x+鈭2/2sin2x)=2鈭2sin(2x+蟺/4)
绛旓細鐪嬪浘瑙ｉ杩囩▼锛
绛旓細涓嶈繃涔熷彲浠ュ彉鎴
绛旓細f(x)=sin²x-cos²x = -cos(2x)鏄伓鍑芥暟锛屾渶灏忔鍛ㄦ湡鏄 D姝ｇ‘
绛旓細=2cos2x+2sin2x =2鈭2sin(2x+蟺/4)瀹炲湪鐪嬩笉鎯竴缇ゅ娓ｇ鏅哄晢缁欎綘涔辫 濡傛灉浣犻棶鏈鍚庝竴姝ユ槸鎬庝箞鎺ㄥ嚭鏉ョ殑 璇疯繍鐢ㄦ櫤鍟嗘兂涓涓嬮珮涓瀛﹁繃鐨勪笁瑙掑嚱鏁拌緟鍔╄鍏紡 acosA+bsinA=鈭(a²+b²)sin(A+M) (tanM=a/b) 铏界劧杩欎釜鍏紡鍙兘鍦ㄩ珮涓敤涓嶈繃浣犲皢灏卞噾鍚堜竴涓嬨傛鐗堝叕寮忎负acosA+b...
绛旓細f(x)=sin2x鈥cos2x=sin2xcos(蟺/4)鈥攃os2xsin(蟺/4)=鈭2sin(2x-蟺/4)鎵浠(x)鍗曡皟閫掑鍖洪棿锛(2k蟺-蟺/2)鈮(2x-蟺/4)<(2k蟺+蟺/2) (k鈭圸)鍗(k蟺-蟺/8)鈮<(k蟺+3蟺/8)锛孠鈭圸 褰搆=0鏃讹紝鍗曞鍖洪棿锛歔-蟺/8,3蟺/8]鈭撮夋嫨B 婊℃剰璇烽噰绾筹紝绁濆涔犺繘姝ワ紒
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(sinx)=sin2x-cos2x,鍥犱负sin锛90+x锛=cosx 鎵浠ュ師寮廸(sin锛坸+90锛)=sin2锛坸+90锛-cos2锛坸+90锛夊緱f锛坈osx锛=-sin2x+cos2x 璇烽噰绾 濡傛灉浣犺鍙垜鐨勫洖绛旓紝鏁鍙婃椂閲囩撼锛寏濡傛灉浣犺鍙垜鐨勫洖绛旓紝璇峰強鏃剁偣鍑汇愰噰绾充负婊℃剰鍥炵瓟銆戞寜閽 ~~鎵嬫満鎻愰棶鐨勬湅鍙嬪湪瀹㈡埛绔彸涓婅璇勪环鐐广愭弧鎰忋戝嵆鍙倊浣...
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