超简单题-------tanx-sinx=(1/cosx-1)是怎么化简的？一步... 求问tanx(1-cosx)是怎么到sinx的
tanx-sinx\u7684sinx\u600e\u4e48\u5316\u62101-cosx\u7684 \u4f9d\u636e\u662f\u4ec0\u4e48 \u6c42\u8bb2\u89e3tanx-sinx
=tanx(1-sinx/tanx)
=tanx(1-sinx/(sinx/cosx)
=tanx(1-cosx)
\u6269\u5c55\u8d44\u6599\uff1a
\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u7684\u57fa\u672c\u5173\u7cfb\u5f0f
\u5012\u6570\u5173\u7cfb\uff1atan\u03b1 \u00b7cot\u03b1=1\u3001sin\u03b1 \u00b7csc\u03b1=1\u3001cos\u03b1 \u00b7sec\u03b1=1\uff1b
\u5546\u7684\u5173\u7cfb\uff1a sin\u03b1/cos\u03b1=tan\u03b1=sec\u03b1/csc\u03b1\u3001cos\u03b1/sin\u03b1=cot\u03b1=csc\u03b1/sec\u03b1\uff1b
\u548c\u7684\u5173\u7cfb\uff1asin^2\u03b1+cos^2\u03b1=1\u30011+tan^2\u03b1=sec^2\u03b1\u30011+cot^2\u03b1=csc^2\u03b1\uff1b
\u5e73\u65b9\u5173\u7cfb\uff1asin²\u03b1+cos²\u03b1=1\u3002
\u7279\u6b8a\u89d2\u7684\u4e09\u89d2\u51fd\u6570\u503c\uff1a
\uff081\uff09sin 0\u00b0 = 0\u3002cos 0\u00b0 = 1\u3001tan 0\u00b0 = 0\u3002
\uff082\uff09sin 30\u00b0 = 1/2\u3001cos 30\u00b0 = \u221a3/2\u3001tan 30\u00b0 = \u221a3/3\u3002
\uff083\uff09sin 45\u00b0 = \u221a2/2\u3001cos 45\u00b0 = \u221a2/2\u3001tan 45\u00b0 = 1\u3002
\uff084\uff09sin 60\u00b0 = \u221a3/2\u3001cos 60\u00b0 = 1/2\u3001tan 60\u00b0 = \u221a3\u3002
\uff085\uff09sin 90\u00b0 = 1\u3001cos 90\u00b0 = 0\u3002
\u8fd9\u662f\u7528\u5230\u4e86\u9ad8\u7b49\u6570\u5b66\u7684\u77e5\u8bc6\uff0c\u7b49\u4ef7\u65e0\u7a77\u5c0f\u3002
\u5c06tanx\uff081-cosx\uff09\u4e0esinx\u6bd4\uff0c\u53ef\u4ee5\u5f97\u52301/cosx-1\uff0c\u5728x\u8d8b\u8fd10\u65f6\uff0c1/cosx-1\u7b49\u4e8e0\u3002
\u53catanx\uff081-cosx\uff09\u4e0esinx\u7b49\u4ef7\u3002
\u4e0d\u61c2\u518d\u8ffd\u95ee\u3002
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tanx-sinx=sinx(1/cosx-1)
tanx=sin/cos
tanx-sinx
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