化简:根号1-cos(2π+a)/1+cos(2π+a) +根号1+cos(2π-a)/1-cos(2π-a)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/11 09:33:56
![化简:根号1-cos(2π+a)/1+cos(2π+a) +根号1+cos(2π-a)/1-cos(2π-a)](/uploads/image/z/2469787-43-7.jpg?t=%E5%8C%96%E7%AE%80%3A%E6%A0%B9%E5%8F%B71-cos%282%CF%80%2Ba%29%2F1%2Bcos%282%CF%80%2Ba%29+%2B%E6%A0%B9%E5%8F%B71%2Bcos%282%CF%80-a%29%2F1-cos%282%CF%80-a%29)
x){3gv>n_atA;QSPQІ*"86IET3KΆ.{g֣YmԌՏ6FjCh V1o=@
`@9}
V*m6\$U:(+dKtjADPLƐj);!U-#Y\Hʂ?
9#=<;hۀ9٧d.`SAq"$RLl25zWI;7?3QYXKjHb3Sv((X)O<[? tDr
化简:根号1-cos(2π+a)/1+cos(2π+a) +根号1+cos(2π-a)/1-cos(2π-a)
化简:根号1-cos(2π+a)/1+cos(2π+a) +根号1+cos(2π-a)/1-cos(2π-a)
化简:根号1-cos(2π+a)/1+cos(2π+a) +根号1+cos(2π-a)/1-cos(2π-a)
化简:√{[1-cos(2π+α)]/[1+os(2π+α)]}+√{[1+cos(2π+α)]/[1-cos(2π-α)]}.
原式=√[(1-cosα)/(1+cosα)]+√[(1+cosα)/(1-cosα)].
=√[(1-cosα)(1+cosα)/(1+cosα)^2]+√[(1+cosα)(1-cosα)/(1-cosα)^2].
=√(1-cos^2α)/(1+cosα)+√(1-cos^2α)/(1-cosα).
=|sinα|/(1+cosαα)+|sinα|/(1-cosα).
=|sinα|[(1-cosα)+(1+cosα)]/(1-cos^2α).
=|sinα|*2/|sinα|.
∴原式=2 .
原式可化为:根号1-c0sa/1+cosa+根号1-cosa/1+cosa
=2倍根号1-ccosa/1+cosa
=2倍根号1-cos平方a/括号1+cosa括号的平方
=2sina/1+cosa
解答完毕