若sinx-cosx=1/3, 则sinx*cosx等于多少，怎么计算 若sinx/2-cosx/2=1/3则sinx等于多少，求详...
sinx+cosx=1/3,\u5219sin2x\u7b49\u4e8e\u591a\u5c11\uff08\u8ba1\u7b97\u8fc7\u7a0b\uff09\uff1f
\u8bf7
\u82e5sinx/2-cosx/2=1/3\u5219sinx\u7b49\u4e8e\u591a\u5c11\uff0c\u6c42\u8be6\u89e3
(sinx/2-cosx/2)²=sin²x/2+cos²x/2-2sinx/2cosx/2=1/9;
2sinx/2cosx/2=1-1/9=8/9;
sinx=8/9;
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(sinx-cosx)^2=sin²x-2sinxcosx+cos²x=1-2sinxcosx=1/9
所以 sinxcosx=4/9
绛旓細鈭sinx-cosx=1锛 鈭磗in 2 x-2sinxcosx+cos 2 x=1锛 鍗1-2sinxcosx=1锛 鈭磗inxcosx=0锛 鏁呯瓟妗堜负锛0锛
绛旓細鈭sinx-cosx=1锛 鈭磗in 2 x-2sinxcosx+cos 2 x=1锛 鍗1-2sinxcosx=1锛 鈭磗inxcosx=0锛 鏁呯瓟妗堜负锛0锛
绛旓細sinx=1+cosx sin²x+cos²x=1 鎵浠(1+cosx)²+cos²x=1锛岃В寰梒osx=0锛宻inx=1鎴朿osx=-1锛宻inx=0 鎵浠inx+cosx=1鎴-1
绛旓細鍘熸柟绋嬪彲鍖栦负锛氭牴鍙2涔樹互sin(x-蟺/4)=1锛宻in(x-蟺/4)=2鍒嗕箣鏍瑰彿2锛屾墍浠-蟺/4=2k蟺+蟺/4鎴杧-蟺/4=2k蟺+3蟺/4 鎵浠=2k蟺+蟺/2鎴杧=2k蟺+蟺 锛坘鏄暣鏁帮級
绛旓細鍛冿紝璇磋鎴戠殑绛旀銆俋-PAI/4=P/4+K*P/2 鎵浠=P/2+K*P/2 涓妤肩瓟妗堟紡浜嗕竴绉嶆儏鍐碉紝涓夋ゼ涓や釜绛旀鏄彲浠ュ悎涓轰竴涓紡瀛愮殑銆備粎渚涘弬鑰冦
绛旓細sinx-cosx=1 鈭2sin(x-蟺/4)=1 sin(x-蟺/4)=鈭2/2 x-蟺/4=蟺/4+2k蟺鎴3蟺/4+2k蟺锛宬鈭圸 瑙ｅ緱 x=蟺/2+2k蟺鎴栂+2k蟺锛宬鈭圸 瑙ｉ泦鏄瘂x|x=蟺/2+2k蟺鎴栂+2k蟺锛宬鈭圸} 鍙傝冭祫鏂欙細http://zhidao.baidu.com/question/419676351.html?quesup2&oldq=1 ...
绛旓細sinx锛峜osx锛1 涓よ竟骞虫柟 鍐嶅埄鐢ㄤ笁瑙掑嚱鏁版ц川鍗虫眰寰 X灞炰簬2鍒嗕箣K娲 娉ㄦ剰k灞炰簬鏁存暟
绛旓細sinx-cosx=-1 鈭2sin(x-蟺/4)=-1 sin(x-蟺/4)=-鈭2/2 x-蟺/4=2k蟺-蟺/4鎴杧-蟺/4=2k蟺-3蟺/4 鎵浠=2k蟺鎴杧=2k蟺-蟺/2,k鏄暣鏁
绛旓細鏍规嵁涓夎鍑芥暟鐨勫畾涔夛紝sin伪=y/r锛宑os伪=x/r锛宺=鈭歺²+y²鎵浠in²伪+cos²伪=1
绛旓細鍘熷紡=(cosx-sinx)^2/[(cosx-sinx)*(cosx+sinx)] 鍊嶈鍏紡 =cosx-sinx 鐢眘inx+cosx=1 鐭ワ細(sinx+cosx)^2=1+2sinxcosx=1=(cosx-sinx)^2+4sinxcosx 鍗砪osx-sinx=1锛-1
