tan（β－α）怎么算是tan(-α＋β)吗？ tan(α-β)怎么算的 还有tan(α+β)
\u5982\u4f55\u7531tan(\u03b1\u00b1\u03b2)\u516c\u5f0f\u63a8\u5bfc\u5230tan\u03b1\u00b1tan\u03b2\u516c\u5f0f\uff1f\uff1f\u9ad8\u624b\u5165\uff01\uff01sin(\u03b1+\u03b2)=sin\u03b1cos\u03b2+cos\u03b1sin\u03b2
cos(\u03b1+\u03b2)=cos\u03b1cos\u03b2-sin\u03b1sin\u03b2
\u4e24\u5f0f\u76f8\u9664
tan(\u03b1+\u03b2)=sin(\u03b1+\u03b2)/cos(\u03b1+\u03b2)=(sin\u03b1cos\u03b2+cos\u03b1sin\u03b2)/(cos\u03b1cos\u03b2-sin\u03b1sin\u03b2)
\u5206\u5b50\u5206\u6bcd\u540c\u9664\u4ee5cos\u03b1cos\u03b2
=(tan\u03b1+tan\u03b2)/(1-tan\u03b1\u00b7tan\u03b2)
tan(\u03b1-\u03b2)=(tan\u03b1-tan\u03b2)/(1+tan\u03b1tan\u03b2)\u3002
tan(\u03b1+\u03b2)=(tan\u03b1+tan\u03b2)/(1-tan\u03b1tan\u03b2)\u3002
\u5206\u6790\u8fc7\u7a0b\u5982\u4e0b\uff1a
tan(\u03b1-\u03b2)\u53ef\u4ee5\u770b\u6210\uff1asin(\u03b1-\u03b2)/cos(\u03b1-\u03b2)
=(sin\u03b1cos\u03b2-cos\u03b1sin\u03b2)/(cos\u03b1cos\u03b2+sin\u03b1sin\u03b2)
\u4e0a\u4e0b\u9664\u4ee5cos\u03b1cos\u03b2
=(tan\u03b1-tan\u03b2)/(1+tan\u03b1tan\u03b2)
tan(\u03b1+\u03b2)\u540c\u7406
=(sin\u03b1cos\u03b2+cos\u03b1sin\u03b2)/(cos\u03b1cos\u03b2-sin\u03b1sin\u03b2)
=(tan\u03b1+tan\u03b2)/(1-tan\u03b1tan\u03b2)
\u6269\u5c55\u8d44\u6599\uff1a
\u79ef\u7684\u5173\u7cfb\uff1a
\uff081\uff09sin\u03b1 = tan\u03b1 \u00d7 cos\u03b1\uff08\u5373sin\u03b1 / cos\u03b1 = tan\u03b1\uff09\u3002
\uff082\uff09cos\u03b1 = cot\u03b1 \u00d7 sin\u03b1 \uff08\u5373cos\u03b1 / sin\u03b1 = cot\u03b1\uff09\u3002
\uff083\uff09tan\u03b1 = sin\u03b1 \u00d7 sec\u03b1 (\u5373 tan\u03b1 / sin\u03b1 = sec\u03b1)\u3002
\u5012\u6570\u5173\u7cfb\uff1a
\uff081\uff09tan\u03b1 \u00d7 cot\u03b1 = 1\u3002
\uff082\uff09sin\u03b1 \u00d7 csc\u03b1 = 1\u3002
\uff083\uff09cos\u03b1 \u00d7 sec\u03b1 = 1\u3002
\u5546\u7684\u5173\u7cfb\uff1a
\uff081\uff09sin\u03b1 / cos\u03b1 = tan\u03b1 = sec\u03b1 / csc\u03b1\u3002
\u548c\u89d2\u516c\u5f0f\uff1a
\uff081\uff09sin ( \u03b1 \u00b1 \u03b2 ) = sin\u03b1 \u00b7 cos\u03b2 \u00b1 cos\u03b1 \u00b7 sin\u03b2\u3002
\uff082\uff09sin ( \u03b1 + \u03b2 + \u03b3 ) = sin\u03b1 \u00b7 cos\u03b2 \u00b7 cos\u03b3 + cos\u03b1 \u00b7 sin\u03b2 \u00b7 cos\u03b3 + cos\u03b1 \u00b7 cos\u03b2 \u00b7 sin\u03b3 - sin\u03b1 \u00b7 sin\u03b2 \u00b7 sin\u03b3\u3002
\uff083\uff09cos ( \u03b1 \u00b1 \u03b2 ) = cos\u03b1 cos\u03b2 ∓ sin\u03b2 sin\u03b1\u3002
\u53c2\u8003\u8d44\u6599\uff1a\u767e\u5ea6\u767e\u79d1-\u6b63\u5f26
tan(-α＋β)=tan(β-α)=(tanβ-tanα)/(1+tanβtanα)
(β－α)=(-α＋β)
绛旓細tan(-伪锛嬑)=tan(尾-伪)=(tan尾-tan伪)/(1+tan尾tan伪)
绛旓細tan(伪-尾)=(tan伪-tan尾)/(1+tan伪tan尾)銆倀an(伪+尾)=(tan伪+tan尾)/(1-tan伪tan尾)銆傚垎鏋愯繃绋嬪涓嬶細tan(伪-尾)鍙互鐪嬫垚锛歴in(伪-尾)/cos(伪-尾)=(sin伪cos尾-cos伪sin尾)/(cos伪cos尾+sin伪sin尾)涓婁笅闄や互cos伪cos尾 =(tan伪-tan尾)/(1+tan伪tan尾)tan(伪+尾)鍚岀悊 ...
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绛旓細tan伪=tan[(伪-尾)+尾]=[tan(伪-尾)+tan尾]/[1-tan(伪-尾)*tan尾]=(5+2)/(1-5*2)=-7/9 涓夎鍏紡瑕佺伒娲昏繍鐢
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