tan(a+b+c)=? tan（a+b）的公式急求
tan(a+b)\u7b49\u4e8e\u4ec0\u4e48tan\uff08a+b\uff09\u7684\u516c\u5f0f\uff1atan(a+b) = (tana+tanb)/(1-tana tanb)\u3002
\u5728Rt\u25b3ABC\uff08\u76f4\u89d2\u4e09\u89d2\u5f62\uff09\u4e2d\uff0c\u2220C=90\u00b0\uff0cAB\u662f\u2220C\u7684\u5bf9\u8fb9c\uff0cBC\u662f\u2220A\u7684\u5bf9\u8fb9a\uff0cAC\u662f\u2220B\u7684\u5bf9\u8fb9b\uff0c\u6b63\u5207\u51fd\u6570tanB=b/a\u3002
\u5728Rt\u25b3ABC\u4e2d\uff0c\u5982\u679c\u9510\u89d2A\u786e\u5b9a\uff0c\u90a3\u4e48\u89d2A\u7684\u5bf9\u8fb9\u4e0e\u90bb\u8fb9\u7684\u6bd4\u503c\u968f\u4e4b\u786e\u5b9a\uff0c\u8fd9\u4e2a\u6bd4\u53eb\u505a\u89d2A\u7684\u6b63\u5207\uff0c\u8bb0\u4f5ctanA\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u4e24\u89d2\u548c\u516c\u5f0f
cot(A+B)=(cotAcotB-1)/(cotB+cotA) cot(A-B)=(cotAcotB+1)/(cotB-cotA)
tan(A+B)=(tanA+tanB)/(1-tanAtanB) tan(A-B)=(tanA-tanB)/(1+tanAtanB)
cos(A+B)=cosAcosB-sinAsinB cos(A-B)=cosAcosB+sinAsinB
sin(A+B)=sinAcosB+cosAsinB sin(A-B)=sinAcosB-sinBcosA
\u79ef\u5316\u548c\u5dee\u516c\u5f0f
sin\u03b1sin\u03b2 = [cos(\u03b1-\u03b2)-cos(\u03b1+\u03b2)] /2
sin\u03b1cos\u03b2 = [sin(\u03b1+\u03b2)+sin(\u03b1-\u03b2)]/2
cos\u03b1sin\u03b2 = [sin(\u03b1+\u03b2)-sin(\u03b1-\u03b2)]/2
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a
\u767e\u5ea6\u767e\u79d1-\u4e09\u89d2\u51fd\u6570\u516c\u5f0f
\u9ad8\u4e2d\u6570\u5b66\uff1a\u5df2\u77e5\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0b\u7684\u4e24\u4e2a\u6839\u4e3atana\u548ctanb\uff0c\u6c42tan\uff08a+b\uff09
第一步:将a+b看做一个角.c看做另外一个角.
tan(a+b+c)=tan[(a+b)+c]=[tan(a+b)+tan(c )]/[1-tan(a+b)*tan(c )]
第二步:再将tan(a+b)展开.所以上式就是.
={{[tan(a)+tan(b)]/[1-tan(a)*tan(b)]}+tan(c )}
/{1-{[tan(a)+tan(b)]/[1-tan(a)*tan(b)]}*tan(c )}
第三步:分子,分母部分都通风,约去[1-tan(a)*tan(b)],整理后得.
=[tan(a)+tan(b)+tan(c )-tan(a)*tan(b)*tan(c )]/[1-tan(a)*tan(b)-tan(a)*tan(c )-tan(b)*tan(c )]
这就是完全展开的结果. (括号比较多,注意分析)
验证一下.
设a+b+c=45度,且a=b=c=15度.我们知道tan(15度)=2-SQRT(3)(即2减根号下3,约等于0.268.)
代入2-SQRT(3),得到tan(45度)=[12*SQRT(3)-20]/[12*SQRT(3)-20]=1.
也可以带如近似值0.268,得到tan(45度)=0.784751168/0.784528 约=1=tan(45度).说明公式正确.
如果是使用单角正切表达的话，可以这样推导：
tan(a + b + c)
= sin(a + b + c)/cos(a + b + c)
= [sina*cos(b + c) + cosa*sin(b + c)] / [cosa*cos(b + c) - sina*sin(b + c)]
分子
= sina*(cosb*cosc - sinb*sinc) + cosa*(sinb*cosc + cosb*sinc)
= sina*cosb*cosc + cosa*sinb*cosc + cosa*cosb*sinc - sina*sinb*sinc
分母
= cosa*(cosb*cosc - sinb*sinc) - sina*(sinb*cosc + cosb*sinc)
= cosa*cosb*cosc - cosa*sinb*sinc - sina*sinb*cosc - sina*cosb*sinc
分子分母同时除以cosa*cosb*cosc（条件是不等于0啊）得到：
tan(a + b + c)
= [tan(a) + tan(b) + tan(c) - tan(a)*tan(b)*tan(c)] / [1 - tan(a)*tan(b) - tan(b)*tan(c) - tan(c)*tan(a)]
tan(a+b+c)=tan[(a+b)+c]
tan[(a+b)+c]=tan(a+b)+tanc/1-tan(a+b)*tanc①
tan(a+b)=tan(a)+tan(b)/1-tan(a)*tan(b)②
②代入①简化后得到如下：
tan(a)+tan(b)+tan(c)-tan(a)*tan(b)*tan(c)/1-tan(a)*tan(b)-tan(b)*tan(c)-tan(c)*tan(a)
/是除,*是乘
Tan(a+b+c)=tan[a+(b+c)]=[tana+tan(b+c)]/tanatan(b+c)
把Tan(b+c)=[tanb+tanc]/tanbtanc代入上式
无解
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绛旓細鍒嗙被: 鏁欒偛/绉戝 >> 瀛︿範甯姪 闂鎻忚堪:鍏蜂綋婕旂畻杩囩▼,璋㈣阿浜嗭紒瑙ｆ瀽:濡傛灉鏄娇鐢ㄥ崟瑙掓鍒囪〃杈剧殑璇濓紝鍙互杩欐牱鎺ㄥ锛tan(a + b + c)= sin(a + b + c)/cos(a + b + c)= [sina*cos(b + c) + cosa*sin(b + c)] / [cosa*cos(b + c) - sina*sin(b + c)]鍒嗗瓙 = sina*(...
