y=cosx-sinx最大值为（） 求y=cosx-sinx的最大值和最小值
\u51fd\u6570y=sinx+cosx\u7684\u6700\u5927\u503c\u4e3a\uff08 \uff09y=sinx+cosx y=\u6839\u53f7\u4e0b2sin(x+\u03c0/4\uff09 \u6240\u4ee5\u5f53x=\u03c0/4\u65f6 \u6709\u6700\u5927\u503c\u4e3a\u6839\u53f72 \u8865\u5145\uff1a sinx\u524d\u7684\u7cfb\u6570\u4e3a1\uff0ccosx\u524d\u7684\u7cfb\u6570\u4e5f\u4e3a1 \u63d0\u51fa\u221a2\uff0c \u6240\u4ee5y=\u221a2\uff08\u221a2/2sinx+\u221a2/2cosx\uff09 cos\u03c0/4=sin\u03c0/4=\u221a2/2 \u6839\u636e\u516c\u5f0fcos\u03b1sin\u03b2+cos\u03b2sin\u03b1=sin\uff08\u03b1+\u03b2\uff09 \u5f97\uff1ay=\u221a2sin(x+\u03c0/4) \u6240\u4ee5x=\u03c0/4\u65f6\uff0cx+\u03c0/4=\u03c0/2\uff0csin(x+\u03c0/4)=1 \u6240\u4ee5\u6700\u5927\u503c\u4e3a\u221a2 \u8865\u5145\uff1a y=\u221a2\uff08\u221a2/2sinx+\u221a2/2cosx\uff09 =\u221a2\uff08cos\u03c0/4sinx+sin\u03c0/4cosx\uff09 =\u221a2sin(x+\u03c0/4) \u8ffd\u95ee\uff1a \u6211\u770b\u4e0d\u61c2......\u545c\u545c......\u80fd\u4e0d\u80fd\u8bf4\u8be6\u7ec6\u70b9\u554a\uff1f\u9ebb\u70e6\u4e86\uff0c\u8c22\u8c22\u3002\u53ef\u80fd\u662f\u6211\u521d\u4e2d\u7684\u77e5\u8bc6\u6ca1\u5b66\u597d\u2026\u2026 \u56de\u7b54\uff1a y=sinx+cosx=\u221a2\uff08\u221a2/2sinx+\u221a2/2cosx\uff09 =\u221a2\uff08cos\u03c0/4sinx+sin\u03c0/4cosx\uff09\uff08\u516c\u5f0f\uff1acos\u03b1sin\u03b2+cos\u03b2sin\u03b1=sin\uff08\u03b1+\u03b2\uff09\uff09 =\u221a2sin(x+\u03c0/4) x+\u03c0/4=\u03c0/2\u65f6\u6700\u5927 sin(x+\u03c0/4)=sin(\u03c0/2)=1 ymax=\u221a2 \u8865\u5145\uff1a \u8fd8\u6709\u4e00\u79cd\u65b9\u6cd5\uff1a \u8bbesinx=y/R,(x^2+y^2=R^2) \u6240\u4ee5cosx=x/R f(x)=y/R+x/R f^2(x)=y^2/R^2+x^2/R^2+2y/R*x/R=1+2y/R*x/R \u6839\u636e\u5747\u503c\u4e0d\u7b49\u5f0f\u5f97\uff1a2y/R*x/R<=y^2/R^2+x^2/R^2=1 f^2(x)<=2 f(x)<=\u221a2 \u6240\u4ee5\u6700\u5927\u503c\u4e3a\u221a2
y=cosx-sinx
y=\u221a2(\u221a2/2cosx-\u221a2/2sinx)
y=\u221a2cos(x+45\u00b0)
y=cosx-sinx\u7684\u6700\u5927\u503c\u221a2,\u6700\u5c0f\u503c-\u221a2\u3002
y=sinx-cosx
y=\u221a2(\u221a2/2sinx-\u221a2/2cosx)
y=\u221a2sin(x-45\u00b0)
y=sinx-cosx\u7684\u6700\u5927\u503c\u221a2,\u6700\u5c0f\u503c-\u221a2\u3002
因为cos（x+π/4）最大值为1
所以y=cosx-sinx最大值为根号二
y=cosx-sinx
= √2(√2/2cosx-√2/2sinx)
= √2cos(x+π/4)
所以最大值是√2
y=cosx-sinx=√2sin(π/4 -x)
所以当sin（(π/4 -x)=1时，取最大值=√2
y=cosx-sinx
=√2(√2/2cosx-√2/2sinx)
=√2cos(x+π/4)
最大值为√2
根号2 我打不出过程
绛旓細鎵浠=cosx-sinx鏈澶у间负鏍瑰彿浜
绛旓細y=cosx-sinx =鈭2脳鈭2/2cosx-鈭2脳鈭2/2sinx =鈭2锛坈os蟺/4cosx-sin蟺/4sinx锛=鈭2cos(x+蟺/4)鏈澶у=鈭2 鏈灏忓=-鈭2 鍛ㄦ湡=2蟺
绛旓細y=cosx-sinx =鏍瑰彿2*(鏍瑰彿2*cosx/2-鏍瑰彿2*sinx/2)=鏍瑰彿2*(cos45掳cosx-sin45掳sinx)=鏍瑰彿2*cos(45掳+x)鎵浠ユ渶澶у=鏍瑰彿2 鏈灏忓=-鏍瑰彿2 绁濅綘瀛︿範杩涙O(鈭鈭)O鍝!
绛旓細y=cosx-sinx鐨勬渶澶у尖垰2,鏈灏忓-鈭2銆倅=sinx-cosx y=鈭2(鈭2/2sinx-鈭2/2cosx)y=鈭2sin(x-45掳)y=sinx-cosx鐨勬渶澶у尖垰2,鏈灏忓-鈭2銆
绛旓細y=cosx-sinx =鈭2[(鈭2/2)cosx-(鈭2/2)sinx]=鈭2cos(x+蟺/4)鎵浠ュ綋cos(x+蟺/4)=1鏃 y鏈夋渶澶у间负鈭2
绛旓細y=cosx-sinx 鈭2
绛旓細鍗冲嚱鏁皔鐨勬瀬澶у间负鈥滄牴2鈥,鏋佸皬鍊间负鈥-鏍2鈥濇柟娉曚簩锛歝osx+sinx=y 鈥︹(1)(cosx)^2+(sinx)^2=1 鈥︹(2)鐢(1)^2-(2)寰 2sinxcosx=y^2-1 鈫抯in2x=y^2-1 鈭-1鈮in2x鈮1 鈭-1鈮^2-1鈮1 瑙ｅ緱,-鏍2鈮鈮ゆ牴2 鍗冲嚱鏁皔鐨勬瀬澶у间负鈥滄牴2鈥,鏋佸皬鍊间负鈥-鏍2鈥.
绛旓細y=鈭2(sinx*鈭2/2-cosx*鈭2/2)=鈭2(sinxcos蟺/4-cosxsin蟺/4)=鈭2sin(x-蟺/4)sin(x-蟺/4)鏈澶=1 鎵浠鏈澶у=鈭2
绛旓細dlw19620101 2017-02-24 路 TA鑾峰緱瓒呰繃2.5涓囦釜璧 鐭ラ亾澶ф湁鍙负绛斾富 鍥炵瓟閲:8911 閲囩撼鐜:57% 甯姪鐨勪汉:2851涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 宸茶禐杩 宸茶俯杩< 浣犲杩欎釜鍥炵瓟鐨勮瘎浠锋槸? 璇勮 鏀惰捣 鍏朵粬绫讳技闂2015-04-02 鍑芥暟y=sinx+cosx鐨鏈澶у间负( ) 30 2016-07-19 姹傚嚱鏁皔=...
绛旓細y=sinx-cosx =鈭2sin(x-蟺/4)鏈澶у=鈭2 姝ゆ椂x-蟺/4=蟺/2+2k蟺,k鏄暣鏁 x=3蟺/4+2k蟺,k鏄暣鏁 {x|x=3蟺/4+2k蟺},k鏄暣鏁 璇峰強鏃剁偣鍑诲彸涓嬭鐨勩愬ソ璇勩戞寜閽
