这个三角函数证明题？？？？？？？？？？、大哥大姐等等都看看(sinO-cscO)(cosO-secO)=1/(tanO+cotO)
\u5df2\u77e5sino+coso=(\u6839\u53f73+1)/2,\u6c42sin/(1-1/tano)+coso/(1-tano)\u7684\u503csino/(1-1/tano)+coso/(1-tano)
=sino/[(sino-coso)/sino]+coso/[(coso-sino)/coso]
=sin²o/(sino-coso)+cos²o/(coso-sino)
=(sin²o-cos²o)/(sino-coso)
=(sino+coso)(sino-coso)/(sino-coso)
=sino+coso
=(\u221a3+1)/2
\u5982\u679c\u60a8\u8ba4\u53ef\u6211\u7684\u56de\u7b54\uff0c\u8bf7\u70b9\u51fb\u201c\u91c7\u7eb3\u4e3a\u6ee1\u610f\u7b54\u6848\u201d,\u795d\u5b66\u4e60\u8fdb\u6b65\uff01
sinx>cosx \u662f(\u03c0/4,5\u03c0/4)\u662f123\u8c61\u9650
cosxsinx\uff1c0 \u662f2 4\u8c61\u9650 \u7efc\u4e0a \u4e3a2\u8c61\u9650
你怎么证明有点看得晕,看我怎么证
左=(sinO-cscO)(cosO-secO)
=(sin0-1/sin0)(cos0-1/cos0)
=sin0cos0-sin0/cos0-cos0/sin0+1/sin0cos0
分开说-sin0/cos0-cos0/sin0合并完是-2/sin20
=sin0cos0-2/sin20+2/sin20=sin0cos0
把分母1展开
=sin0cos0/(sin0^2+cos0^2)
=1/(tan0+cot0)
得证
问老师去
(*^__^*) 嘻嘻……我的意思是你右边式子的变形错了。。。左边式子的变形正确。。
1/(tanO+cotO)变形后应该为sinO*cosO因此你需证明-tanO-cotO+1/(sinO*cosO)等于0就可以了哈。。
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