X∈(0,π/2),sinαcosα=1/2,求1/(1+sinα)+1/(1+cosα).(π就是派)答案是负的(1+2倍根号2)/3。老师提示是通分
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![X∈(0,π/2),sinαcosα=1/2,求1/(1+sinα)+1/(1+cosα).(π就是派)答案是负的(1+2倍根号2)/3。老师提示是通分](/uploads/image/z/227251-19-1.jpg?t=X%E2%88%88%280%2C%CF%80%2F2%29%2Csin%CE%B1cos%CE%B1%3D1%2F2%2C%E6%B1%821%2F%EF%BC%881%2Bsin%CE%B1%EF%BC%89%2B1%2F%281%2Bcos%CE%B1%29.%EF%BC%88%CF%80%E5%B0%B1%E6%98%AF%E6%B4%BE%EF%BC%89%E7%AD%94%E6%A1%88%E6%98%AF%E8%B4%9F%E7%9A%84%EF%BC%881%2B2%E5%80%8D%E6%A0%B9%E5%8F%B72%EF%BC%89%2F3%E3%80%82%E8%80%81%E5%B8%88%E6%8F%90%E7%A4%BA%E6%98%AF%E9%80%9A%E5%88%86)
X∈(0,π/2),sinαcosα=1/2,求1/(1+sinα)+1/(1+cosα).(π就是派)答案是负的(1+2倍根号2)/3。老师提示是通分
X∈(0,π/2),sinαcosα=1/2,求1/(1+sinα)+1/(1+cosα).(π就是派)
答案是负的(1+2倍根号2)/3。老师提示是通分
X∈(0,π/2),sinαcosα=1/2,求1/(1+sinα)+1/(1+cosα).(π就是派)答案是负的(1+2倍根号2)/3。老师提示是通分
1/(1+sinα)+1/(1+cosα)
=(1+cos+1+sin)/(1+sin)(1+cos)
=(2+sin+cos)/(1+sin+cos+sinαcosα)
∵sinαcosα=1/2,sinα ×sinα +cosα×cosα=1,
∴(sin+cos)×(sin+cos)=1+1=2.
又∵X∈(0,π/2),
∴sin+cos=根号2,
∴原式=(2+根号2)/(1+根号2+1/2)=4-2×根号2
cos^2(α/2)=(1+cosα)/2和sin^2(α/2)=(1-cosα)/2,sinαcosα=1/2sin2α是解题关键
=4-2根号2
sinαcosα=1/2 sinα平方+cosα平方=1 解方程
最后求得sinα=(根号2)/ 2 cosα=(根号2)/ 2
然后代入求。。。分母有理化。。。一直求就有答案了
因为sinαcosα=1/2,所以sina2α=1,即2α=kπ+π/2,所以α=kπ/2+π/4,又X∈(0,π/2),所以α=π/4
1/(1+sinα)+1/(1+cosα)。。。=n/(1+sinα)=n/(1+sinaπ/4)=2n/(2+根号2)
sinαcosα=1/2,所以sinα+cosα=sqr(1+2*1/2)=sqr(2)
因此1/(1+sinα)+1/(1+cosα)=(2+sinα+cosα)/(1+sinα+cosα+sinαcosα)=(2+根号2)/(1+根号2+1/2)=4-2sqr(2)