sin2x-cos2x怎么化简？

计算过程如下：

sin2X-cos2X

=√2sin2Xcosπ/4-√2cos2Xsinπ/4

=√2sin（2x-π/4）

倍角半角公式：

sin ( 2α ) = 2sinα · cosα

sin ( 3α ) = 3sinα - 4sin & sup3 ; ( α ) = 4sinα · sin ( 60 + α ) sin ( 60 - α )

sin ( α / 2 ) = ± √( ( 1 - cosα ) / 2)

由泰勒级数得出

sinx = [ e ^ ( ix ) - e ^ ( - ix ) ] / ( 2i )

sin2x-cos2x鎬庝箞鍖栫畝?
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...

y=sin2x-cos2x鎬庝箞鍖栫畝
绛旓細=sin2x-cos2x =鈭2.sin(2x-蟺/4)

sin2x-cos2x鎬庝箞鍖栫畝
绛旓細瑙鍖栫畝鏄牴2sin(x-鍏/4)

鍑芥暟f(x)=sin2x-cos2x,鎬庝箞鍖栫畝鍒癴(x)=鈭2sin(2x-蟺/4)鐨..姹傝В
绛旓細=鈭2*[sin2x*cos(蟺/4) -cos2x*sin(蟺/4)]=鈭2*sin(2x-蟺/4)

sin2x-cos2x鐨鍖栫畝涓嶈兘涓嶈兘鎶2鍑烘潵,鐒跺悗鍙樻垚:2(½sin2x-½cos...
绛旓細浼间箮涓嶈兘,濡備笅 涓嶈繃涔熷彲浠ュ彉鎴

瑙ｅ喅y=sin2x-cos2x鐨勫鏁版槸?瑕佹纭殑姝ラ
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)

cos2x-sin2x鎬庝箞鍖栫畝
绛旓細cos2x-sin2x =鈭2*(cos2x*鈭2/2 - sin2x*鈭2/2)=鈭2*[cos2x*cos(蟺/4) - sin2x*sin(蟺/4)]=鈭2*cos(2x+ 蟺/4)

cos2x-sin2x鍙互鐩存帴鐢ㄨ緟鍔╄鍏紡鍚 濡傛灉涓嶈,搴旇鎬庝箞鍖栫畝
绛旓細cos2x-sin2x=cosx^2-sinx^2-2sinxcosx =(cosx-sinx)^2-2sinx^2 =(cosx+(-1+2^0.5)sinx)(cosx-(1+2^0.5)sinx)

sin2xcos2x鎬庝箞鍖栫畝?
绛旓細浣犳槸瑕鍖栫畝鎴愪粈涔堟牱鐨勫晩 y=sin2xcos2x=1/2sin4x锛岃繖涓彲浠ヤ笉锛熻閲囩撼绛旀锛屾敮鎸佹垜涓涓嬨

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