cos3x怎样变成cosx? cosx－cos3x得多少？有具体步骤不？
\u5982\u56fe cos3xcosx\u662f\u5982\u4f55\u53d8\u6362\u7684\uff1fcos4x+cos2x
=cos(3x+x)+cos(3x-x)
=cos3xcosx-sin3xsinx+cos3xcosx+sin3xsinx
=2cos3xcosx.
\u5012\u8fc7\u6765\u5c31\u884c
\u671b\u91c7\u7eb3 \u8c22\u8c22
\u4e0d\u61c2\u5f97\u95ee\u6211
\u5177\u4f53\u56de\u7b54\u5982\u4e0b\uff1a
cosx-cos3x
=cos(2x-x)-cos(2x+x)
=(cos2xcosx+sin2xsinx)-(cos2xcosx-sin2xsinx)
=2sin2xsinx
\u51fd\u6570\u7684\u7b97\u5f0f\uff1a
\u79ef\u7684\u5173\u7cfb\uff1a
sin\u03b1 = tan\u03b1 \u00d7 cos\u03b1\uff08\u5373sin\u03b1 / cos\u03b1 = tan\u03b1 \uff09
cos\u03b1 = cot\u03b1 \u00d7 sin\u03b1 \uff08\u5373cos\u03b1 / sin\u03b1 = cot\u03b1\uff09
tan\u03b1 = sin\u03b1 \u00d7 sec\u03b1 (\u5373 tan\u03b1 / sin\u03b1 = sec\u03b1)
\u5012\u6570\u5173\u7cfb\uff1a
tan\u03b1 \u00d7 cot\u03b1 = 1
sin\u03b1 \u00d7 csc\u03b1 = 1
cos\u03b1 \u00d7 sec\u03b1 = 1
cos3x变成cosx,w变成1/3，则周期要变为原来三倍，即拉伸3倍。
cos(3α) = 4cos^3α-3cosα = 4cosα·cos(π/3+α)cos(π/3-α)
cos3a
=cos(2a+a)
=cos2acosa-sin2asina
=(2cos²a-1)cosa-2(1-cos²a)cosa
=4cos³a-3cosa
cos3a=4cos³a-3cosa
=4cosa(cos²a-3/4)
=4cosa[cos²a-(√3/2)²]
=4cosa(cos²a-cos²30°)
=4cosa(cosa+cos30°)(cosa-cos30°)
=4cosa*2cos[(a+30°)/2]cos[(a-30°)/2]*{-2sin[(a+30°)/2]sin[(a-30°)/2]}
=-4cosasin(a+30°)sin(a-30°)
=-4cosasin[90°-(60°-a)]sin[-90°+(60°+a)]
=-4cosacos(60°-a)[-cos(60°+a)]
=4cosacos(60°-a)cos(60°+a)
cos(x+2x)=cosx^2-sin2x^2
再把cos2x用和角公式分开就行了
绛旓細瑕佹眰cos³x锛屽繀椤诲緱鍒cosx 浜庢槸杩欎釜涓夊嶈鍏紡鍙樻垚涓涓叧浜巆osx鐨勪笁娆℃柟绋4cos³x锕3cosx锕cos3x=0鈥斺斺憼 鍏朵腑cos3x涓哄凡鐭ラ噺銆傛柟绋嬧憼鐨勮В蹇呴』杩涘叆澶嶆暟鍩熸墠鑳藉緱鍒拌В绛斻傜敤鍗″皵涓瑰叕寮忚В鏂圭▼鈶狅紝寰楀埌涓変釜鐢ㄥ鏁拌〃绀虹殑瀹炴暟鏍癸紝寰堝鏉傦紝鎴戜笉鍐欎簡锛岃嚜宸辩櫨搴﹀惂銆備篃灏辨槸璇村凡鐭ヤ换鎰忎竴涓猻in3x鎴朿os...
绛旓細f(cosx)=cos3x=cos(x+2x)=cosxcos2x-sinxsin2x =cosx(2cos²x-1)-2sin²xcosx =2cos³x-cosx-2(1-cos²x)cosx =2cos³x-cosx-2(cosx-cos³x)=4cos³x-3cosx 浠osx涓簒.鎵浠 f(x)=4x³-3x ...
绛旓細鍏堝埄鐢ㄥ拰鐨勪綑寮﹀叕寮忥紝cos3a=cos(2a+a)=cos2acosa-sin2asina锛屽啀鍒╃敤鍊嶈鍏紡锛=(2cos^2-1)cos-2sin^2cos.鍒嗛厤寰嬪苟鍒╃敤姝ｅ鸡鍜屼綑寮︾殑骞虫柟鍜屽叕寮忥紝=2cos^3-cos-(1-cos^2)cos 涔樻硶鍒嗛厤寰嬪悎骞跺悓绫婚」灏卞緱鍒版渶鍚庣殑缁撴灉浜
绛旓細鍙樺舰寰楀埌锛cos3x(2cosx-1)=0 cos3x=0 鎴栬 2cosx=1鍗 cosx=1/2 3x=蟺/2 锛屾垨 x=蟺/3 x=蟺/6 锛屾垨 x=蟺/3 ~浜诧紝濡傛灉浣犺鍙垜鐨勫洖绛旓紝璇风偣鍑汇愰噰绾充负婊℃剰鍥炵瓟銆戞寜閽畘~鎵嬫満鎻愰棶鐨勬湅鍙嬪湪瀹㈡埛绔笂璇勪环鐐广愭弧鎰忋戝嵆鍙倊浣犵殑閲囩撼鏄垜鍓嶈繘鐨勫姩鍔泘~O(鈭鈭)O锛屼簰鐩稿府鍔╋紝绁濆叡鍚岃繘姝ワ紒
绛旓細cos2x=2(cosx)^2-1,cos2x=2锛坰inx锛塣2+1,con2x=(cosx)^2-锛坰inx锛塣2,tan2x=2tanx/[1-(tanx)^2],sin3x=3sinx-4(sinx)^3 ,sin4x=2sin2x-2sin2xcos2x,cos3x=4(cos)^3-3cosx,sin2x=2sinxcosx
绛旓細cos3x=cos(2x+x)=cos2xcosx-sin2xsinx =(2cosxcosx-1)cosx-2(1-cosxcosx)cosx 甯﹀叆鍗冲彲
绛旓細鍙樺舰寰楀埌锛cos3x(2cosx-1)=0 cos3x=0 鎴栬 2cosx=1鍗 cosx=1/2 3x=蟺/2 锛屾垨 x=蟺/3 x=蟺/6 锛屾垨 x=蟺/3 ~浜诧紝濡傛灉浣犺鍙垜鐨勫洖绛旓紝璇风偣鍑汇愰噰绾充负婊℃剰鍥炵瓟銆戞寜閽畘~鎵嬫満鎻愰棶鐨勬湅鍙嬪湪瀹㈡埛绔笂璇勪环鐐广愭弧鎰忋戝嵆鍙倊浣犵殑閲囩撼鏄垜鍓嶈繘鐨勫姩鍔泘~O(鈭鈭)O锛屼簰鐩稿府鍔╋紝绁濆叡鍚岃繘姝ワ紒
绛旓細鍥犱负cos(a+b)+cos(a-b)=2cosacosb(鍜屽樊鍖栫Н鍏紡锛変护x=a-b,3x=a+b 閭ｄ箞鍘熷紡鍙敼鍐欎负锛cosx+cos3x=2(cos(x+3x)/2)*锛坈os(3x-x)/2)=2cos2x*cosx
绛旓細x)=2sinx(1-sin²x)+sinx(1-2sin²x)=3sinx-4sin³x寰楄瘉 鈶¤姹傚嚭P4锛坱锛夛紝鍗崇敤涓涓cosx鐨勫洓娆″椤瑰紡鏉ヨ〃绀篶os4x 涓ゆ杩愮敤鍊嶈鍏紡cos2x=2cos²x-1锛屽彲鐭os4x=2cos²(2x)-1=2(2cos²x-1)^2-1=8(cosx)^4-8cos²x+1 ...
绛旓細cos3x=cos(4x-x)=cos4xcosx+sin4xsinx,cos5x=cos(4x+x)=cos4xcosx-sin4xsinx,鑰岋紝cos3x=cos5x,鎵浠ユ湁cos4xcosx+sin4xsinx=cos4xcosx-sin4xsinx.寰楀埌2sin4xsinx=0,浠庤屾湁x=kpai,鎴4x=kpai.
