1-cos2x等于2sinx^2求详细推导过程 求高手证明Sinx的平方为什么等于(1-Cos2x)/2

1-cos2x\u7b49\u4ef7\u65e0\u7a77\u5c0f\u662f(2x²)/2\uff0c\u8bf7\u95ee\u600e\u4e48\u63a8\u5bfc\u51fa\u6765\u7684\uff0c\u8fc7\u7a0b\u5199\u4e00\u4e0b

1-cos2x\uff1d1-\uff08cosx\uff09^2+(sinx)^2\uff1d1-[1-\uff08sinx\uff09^2]+(sinx)^2\uff1d2(sinx)^2
\u56e0\u4e3asinx~x \u540c\u65f6\u5e73\u65b9sin^2x~x^2 \u800csin^2x\u7b49\u4e8e\uff081-cos2x\uff09/2
\u6545\uff081-cos2x\uff09/2~x^2 \u6240\u4ee51-cos2x~2x^2
\u518d\u5c06x=2t\u5e26\u5165\u5f971-cost~t^2/2

\u6269\u5c55\u8d44\u6599\uff1a
\u6709\u9650\u4e2a\u65e0\u7a77\u5c0f\u91cf\u4e4b\u548c\u4ecd\u662f\u65e0\u7a77\u5c0f\u91cf\u3002
\u6709\u9650\u4e2a\u65e0\u7a77\u5c0f\u91cf\u4e4b\u79ef\u4ecd\u662f\u65e0\u7a77\u5c0f\u91cf\u3002
\u6709\u754c\u51fd\u6570\u4e0e\u65e0\u7a77\u5c0f\u91cf\u4e4b\u79ef\u4e3a\u65e0\u7a77\u5c0f\u91cf\u3002
\u7279\u522b\u5730\uff0c\u5e38\u6570\u548c\u65e0\u7a77\u5c0f\u91cf\u7684\u4e58\u79ef\u4e5f\u4e3a\u65e0\u7a77\u5c0f\u91cf\u3002
\u6052\u4e0d\u4e3a\u96f6\u7684\u65e0\u7a77\u5c0f\u91cf\u7684\u5012\u6570\u4e3a\u65e0\u7a77\u5927\uff0c\u65e0\u7a77\u5927\u7684\u5012\u6570\u4e3a\u65e0\u7a77\u5c0f\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u65e0\u7a77\u5c0f\u91cf

\u8fd9\u662f\u548c\u89d2\u516c\u5f0f\u7684\u53d8\u5f62
cos(A+B)=cosAcosB-sinAsinB
\u5f53A=B=x\u7684\u65f6\u5019\uff0c\u4f60\u4ee3\u5165\u7b97\u7b97\uff0ccos(2x)=(cosx)^2-(sinx)^2
\u518d\u7ed3\u5408(cosx)^2+(sinx)^2=1
\u4e24\u5f0f\u8054\u7acb\uff0c\u5c31\u5f97\u51fa\u4f60\u8981\u7684\u7ed3\u679c\u4e86

1-cos2x＝1-（cosx）^2+(sinx)^2＝1-[1-（sinx）^2]+(sinx)^2＝2(sinx)^2

原式=1-cos(x+x)=1-cos^2x+sin^2x=2sin^2x

1鈥cos2X 绛変簬澶氬皯
绛旓細鍥炵瓟锛氬師寮=1-(1-2sin²x) =2sin²x

1-cos2x绛変簬澶氬皯?
绛旓細1-cos2x=1-[1-2*(sinx)^2]=2脳(sinx)^2銆傚叾浠栨嫇灞曞叕寮 cos2x=cos²x-sin²x cos2x=1-2sin²x cos2x=2cos²x-1 cos2x=1-(1-2sin²x)=2sin²x cos3a =cos(2a+a)=cos2acosa-sin^2asina =(2cos^2a-1)cosa-2(1-cos^2a)cosa =4cos^3a-...

1鍑忓幓cos2x绛変簬浠涔
绛旓細2脳(sinx)^2銆傜敱鍏紡锛氬彲寰楋細cos2x=2*(cosx)^2-1=1-2*(sinx)^2 鏁咃細1-cos2x=1-[1-2*(sinx)^2]=2脳(sinx)^2銆傚嶈鍏紡锛屾槸涓夎鍑芥暟涓潪甯稿疄鐢ㄧ殑涓绫诲叕寮忋傚氨鏄妸浜屽嶈鐨勪笁瑙掑嚱鏁扮敤鏈鐨勪笁瑙掑嚱鏁拌〃绀哄嚭鏉ャ傚湪璁＄畻涓彲浠ョ敤鏉ュ寲绠璁＄畻寮忋佸噺灏戞眰涓夎鍑芥暟鐨勬鏁帮紝鍦ㄥ伐绋嬩腑涔熸湁骞挎硾鐨勮繍...

1-cos2x绛変簬2sinx^2姹傝缁嗘帹瀵艰繃绋
绛旓細1-cos2x锛1-锛坈osx锛^2+(sinx)^2锛1-[1-锛坰inx锛塣2]+(sinx)^2锛2(sinx)^2

1- cos2x绛変簬澶氬皯?绛変环鏃犵┓灏忔槸浠涔?
绛旓細cos2x=1-2sinx^2銆傛墍浠1-cos2x=2sinx^2銆傚綋x瓒嬩簬0鏃讹紝sinx锝瀤銆傛墍浠瓒嬩簬0鏃讹紝sinx^2锝瀤^2銆傛墍浠1鈥攃os2x绛変环鏃犵┓灏忔槸2x^2銆備簩鍊嶈鍏紡鐨勮繍鐢 浜屽嶈鍏紡閫氳繃瑙捨辩殑涓夎鍑芥暟鍊肩殑涓浜涘彉鎹㈠叧绯绘潵琛ㄧず鍏朵簩鍊嶈2伪鐨勪笁瑙掑嚱鏁板硷紝浜屽嶈鍏紡鍖呮嫭姝ｅ鸡浜屽嶈鍏紡銆佷綑寮︿簩鍊嶈鍏紡浠ュ強姝ｅ垏浜屽嶈...

鎬庢牱璁＄畻1- cos(2x)?
绛旓細浜屽嶈鍏紡锛歝os(2x)=(cosx)^2-(sinx)^2 =2(cosx)^2-1 =1-2(sinx)^2 鍒欏彲寰楋細1- cos(2x)=1-[1-2(sinx)^2]=2(sinx)^2

1-cos2x涓2sinx^2鏄浉鍚屽嚱鏁板悧?
绛旓細1 - cos2x 锛 2sin²x锛岃繖鏄鎭掔瓑寮忥紝涓よ竟瀹屽叏鐩稿悓銆

楂樻墜```1-cosx鎬庝箞绛変簬2鍊嶇殑(sin x/2)^2 鐢ㄤ簡浠涔堝叕寮 鎬庝箞鎺ㄧ殑_鐧 ...
绛旓細鍥犱负cos2x=(cosx)^2 -(sinx)^2 鑰屼笖1=(cosx)^2 +(sinx)^2 鎵浠1-cos2x=2(sinx)^2 鍥犳 1-cosx=2(sinx/2)^2

宸茬煡f(sinx)=1-cos2x,姹俧(cosx)绛変簬澶氬皯,瑕佽繃绋,璋㈣阿!!
绛旓細f锛sinx锛=1-cos2x=2(sin x)^2,鏃㈡湁f(x)=2x^2,鏁協(cos x)=2*(cos x)^2=1+cos2x.

1-cos2x绛変环鏃犵┓灏忔槸(2x²)/2,璇烽棶鎬庝箞鎺ㄥ鍑烘潵鐨,杩囩▼鍐欎竴涓
绛旓細1-cos2x锛1-锛坈osx锛塣2+(sinx)^2锛1-[1-锛坰inx锛塣2]+(sinx)^2锛2(sinx)^2 鍥犱负sinx~x 鍚屾椂骞虫柟sin^2x~x^2 鑰宻in^2x绛変簬锛1-cos2x锛/2 鏁咃紙1-cos2x锛/2~x^2 鎵浠1-cos2x~2x^2 鍐嶅皢x=2t甯﹀叆寰1-cost~t^2/2

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