y=sin2xcos2x怎么化成最简单的式子？ y=sin2xcos2x的最小正周期是多少 能告诉我最简单的...

\u51fd\u6570y=sin2xcos2x\u600e\u4e48\u5316\u7b80

(1/2)*sin4x
\u89e3\uff1a
y
=sin2xcos2x
=(1/2)*2sin2xcos2x
=(1/2)*sin4x
PS\uff1a\u4ea6\u53ef\u4ee5\u5316\u4e3a\u5176\u5b83\u5f62\u5f0f

\u539f\u5f0f=1/2sin4x \u6240\u4ee5\u6700\u5c0f\u6b63\u5468\u671f\u4e3a\u03c0/2\u5c31\u662f\u4e66\u4e0a\u516c\u5f0f\u7684\u5e94\u7528\uff1asinxcosx=1/2sin2x

答案：y=1/2sin4x
呐，为了方便，我们不妨令2x=A
那么y=sinAcosA
=1/2*2sinAcosA
=1/2sin2A(两角将公式）
=1/2sin4x

厄，懂否？楼上好像错了。

y=1/2*2sin2xcos2x
y=1/2*sin4x

y=1/2*2*sin2xcos22x=1/2sin4x

考查的是二倍角公式，sin2a=2sinacosa

y=sin4x/2

答案是1/2sin4x, 我刚好有看到一道一样的选择题哈

y=sin2xcos2x鎬庝箞鍖栫畝?
绛旓細y=sin(2x)cos(2x)=(1/2)sin(4x)

y=sin²x.cos2x鎬庝箞鍖栬В?
绛旓細y=sin^2(x)cos2x=(1-cos2x)/2*cos2x=(cos2x)/2-(cos2x)^2/2=(cos2x)/2-(1 cos4x)/4 y'= -sin2x sin4x

y=sin2xcos2x鎬庝箞鍖栫畝
绛旓細鏍规嵁2鍊嶈鍏紡sin(2x)=2sinxcosx锛屽緱 y=sin2xcos2x=1/2(2sin2xcos2x)=1/2sin4x

鍑芥暟y=sin2xcos2x鎬庝箞鍖栫畝
绛旓細瑙ｏ細y =sin2xcos2x =(1/2)*2sin2xcos2x =(1/2)*sin4x PS锛氫害鍙互鍖栦负鍏跺畠褰㈠紡

y=sin2xcos2x鍖栫畝鎬庝箞鍒皔=sin4x/2鐨,瑕佸叿浣撴楠
绛旓細涓夎鍑芥暟鍏紡sin2t=2sintcost (sin2t)/2=sintcost 浠2x=t y=sin2xcos2x=sintcost=(sin2t)/2 2t=4x 鎵浠 y=sin2xcos2x=锛坰in4x锛/2

y=sin2xcos2x鍖栫畝鎬庝箞鍒皔=sin4x/2鐨,
绛旓細涓夎鍑芥暟鍏紡sin2t=2sintcost (sin2t)/2=sintcost 浠2x=t y=sin2xcos2x=sintcost=(sin2t)/2 2t=4x 鎵浠 y=sin2xcos2x=锛坰in4x锛/2

y=sin2xcos2x鎬庝箞鍖栨垚鏈绠鍗曠殑寮忓瓙?
绛旓細绛旀锛歽=1/2sin4x 鍛愶紝涓轰簡鏂逛究锛屾垜浠笉濡ㄤ护2x=A 閭ｄ箞y=sinAcosA =1/2*2sinAcosA =1/2sin2A(涓よ灏嗗叕寮忥級=1/2sin4x 鍘勶紝鎳傚惁锛熸ゼ涓婂ソ鍍忛敊浜嗐

鍑芥暟y=sin2xcos2x鎬庝箞鍖栨垚1/2 sin4x,姹傝繃绋
绛旓細鏍规嵁2鍊嶈鍏紡锛歴in(2x)=2sinxcosx y=sin2xcos2x=1/2(2sin2xcos2x)=1/2sin4x

y=sin2xcos2x=1/2sin4x鎬庝箞鍋氱殑姹傝缁嗚繃绋
绛旓細瑙ｏ細宸茬煡锛y=sin2xcos2x 鍙樻崲锛歽=(1/2)(2sin2xcos2x)鐢卞嶈鍏紡锛歽=(1/2)sin(4x)

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