sin2x-cos^2x怎么化简
2sinxcosx绛旓細浣犲ソ锛氫负浣犳彁渚涜缁嗚В绛 y=sin^2x-cos^2x=-cos2x 鎵浠ュ懆鏈熶负2蟺/2=蟺 鏍规嵁浣欏鸡鍑芥暟鐨勬湁鐣屾э紝鏈灏忓间负-1 璋㈣阿涓嶆噦鍙拷闂
绛旓細sin^2x-cos^2x =1-cos^2x-cos^2x =1-2cos^2x 鍛ㄦ湡T=2蟺/2=蟺 褰揷os^2x=0鏃跺彇寰楁渶澶у1 鍝︽垜鐭ラ亾浜唦鍥犱负鏄簩娆＄殑浣欏鸡鍑芥暟鎵浠os^x鐨勫懆鏈熷簲璇ユ槸蟺锛堢敾鍥惧彲浠ョ湅鍑烘潵鐨勶紝浜屾鐨勮瘽鎵鏈夊奸兘鏄潪璐熸暟锛夛紝鎵浠ュ懆鏈烼=蟺/2 涓嶅ソ鎰忔濆垰鎵嶆病鐪嬫竻妤殈璇浜 ...
绛旓細y=sin^2x-cos^2x =-(cos^2x-sin^2x)=-cos2x -1<=cos2x<=1 -1<=-cos2x<=1 鎵浠ュ嚱鏁皔=sin^2x-cos^2x鐨勬渶灏忓兼槸锛-1
绛旓細浠(x)=sinx*sinx-cosx*cosx=-cos2x g(x)=cosx鏄伓鍑芥暟 鎵浠ュ師鍑芥暟鏄伓鍑芥暟 f(x)鐨勫懆鏈熸槸蟺锛屽湪(k蟺锛宬蟺+蟺/2)鍐呭崟璋冮掑锛屽湪(k蟺+蟺/2锛宬蟺+蟺)鍐呭崟璋冮掑噺銆
绛旓細y= 3 sin2x-cos2x=2(3 2 sin2x- 1 2 cos2x)=2sin(2x- 蟺 6 )=2sin2(x- 蟺 12 )锛庢牴鎹乏鍔犲彸鍑忕殑鍘熷垯锛岃寰楀埌鍑芥暟y=2sin2x鐨勫浘璞″彧瑕佸皢y= 3 sin2x-cos2x鐨勫浘璞″悜宸﹀钩绉 蟺 12 涓崟浣
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)
绛旓細=sin2x-(1+cos2x)=sin2x-cos2x-1 =鈭2(鈭2/2sin2x-鈭2/2cos2x)-1 =鈭2sin(2x-蟺/4)-1;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犵殑閲囩撼鏄垜鍓嶈繘鐨勫姩鍔 ~~O(鈭...
绛旓細cos2x-sin2x =鈭2脳[cos2x*(鈭2/2)-sin2x*(鈭2/2)]=鈭2脳[cos2x*cos(蟺/4)-sin2x*sin(蟺/4)]=鈭2cos[2x+(蟺/4)]
绛旓細杩欐槸鍜岃鍏紡 鍘熷紡=cos(2x)
绛旓細sin2x-cos2x-1 =鈭2(sin2x脳鈭2/2-cos2x脳鈭2/2)-1 =鈭2sin(2x-蟺/4)-1 濡傛灉鏈鏈変粈涔堜笉鏄庣櫧鍙互杩介棶锛屽鏋滄弧鎰忚寰楅噰绾 濡傛灉鏈夊叾浠栭棶棰樿閲囩撼鏈鍚庡彟鍙戠偣鍑诲悜鎴戞眰鍔╋紝绛旈涓嶆槗锛岃璋呰В锛岃阿璋傜瀛︿範杩涙
