已知tanα=2,α∈(π,3π/2)(1)求sin(π+α)+2sin(3π/2+α) / cos(3π-α)+1 (2)√3cosα-sinα / √3cosα+sinα第一问的1化简不掉怎么办
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![已知tanα=2,α∈(π,3π/2)(1)求sin(π+α)+2sin(3π/2+α) / cos(3π-α)+1 (2)√3cosα-sinα / √3cosα+sinα第一问的1化简不掉怎么办](/uploads/image/z/5189759-71-9.jpg?t=%E5%B7%B2%E7%9F%A5tan%CE%B1%3D2%2C%CE%B1%E2%88%88%28%CF%80%2C3%CF%80%2F2%29%281%29%E6%B1%82sin%28%CF%80%2B%CE%B1%29%2B2sin%283%CF%80%2F2%2B%CE%B1%29+%2F+cos%283%CF%80-%CE%B1%29%2B1+%282%29%E2%88%9A3cos%CE%B1-sin%CE%B1+%2F+%E2%88%9A3cos%CE%B1%2Bsin%CE%B1%E7%AC%AC%E4%B8%80%E9%97%AE%E7%9A%841%E5%8C%96%E7%AE%80%E4%B8%8D%E6%8E%89%E6%80%8E%E4%B9%88%E5%8A%9E)
已知tanα=2,α∈(π,3π/2)(1)求sin(π+α)+2sin(3π/2+α) / cos(3π-α)+1 (2)√3cosα-sinα / √3cosα+sinα第一问的1化简不掉怎么办
已知tanα=2,α∈(π,3π/2)
(1)求sin(π+α)+2sin(3π/2+α) / cos(3π-α)+1
(2)√3cosα-sinα / √3cosα+sinα
第一问的1化简不掉怎么办
已知tanα=2,α∈(π,3π/2)(1)求sin(π+α)+2sin(3π/2+α) / cos(3π-α)+1 (2)√3cosα-sinα / √3cosα+sinα第一问的1化简不掉怎么办
(1)
原式=(-sina-2cosa)/(-cosa+1)
=(sina+2cosa)/(cosa-1)
1无法消掉,
只能求出sina,cosa
tana=2,a是第三象限角
∴ sina=-2/√5,cosa=-1/√5
代入即可
(2)
分子分母同时除以cosa
原式=(√3-tana)/(√3+tana)
=(√3-2)/(√3+2)
=-(√3-2)²
=-7+4√3
1应该不是在分母的 (1) sin(π+α)+2sin(3π/2+α)/cos(3π-α) +1 =(-sinα-2cosα)/(-cosα)+1 =tanα+2+1 (2)(√3cosα-sinα)/(√3cosα+sinα) =(√3cosα/cosα-sinα/cosα)/(√3cosα/cosα+sinα/cosα) (分子分母同时除以cosa) =(√3-tanα)/(√3+tanα) =(√3-2)/(√3+2) =(√3-2)(2-√3) /(√3+2) (2-√3) =-(2-√3)²/(4-3) =4√3-7
=[(sina/cosa)+2]+1 (分子分母同时除以-cosa得)
=2+2+1
=5