已知sinα,cosα是方程2x²-x+a=0的两个根,求下列各值sinαcosα1/sinα+1/cosαsin四次方α+cos四次方αsin³α+cos³α
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![已知sinα,cosα是方程2x²-x+a=0的两个根,求下列各值sinαcosα1/sinα+1/cosαsin四次方α+cos四次方αsin³α+cos³α](/uploads/image/z/15192251-35-1.jpg?t=%E5%B7%B2%E7%9F%A5sin%CE%B1%2Ccos%CE%B1%E6%98%AF%E6%96%B9%E7%A8%8B2x%26%23178%3B-x%2Ba%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%2C%E6%B1%82%E4%B8%8B%E5%88%97%E5%90%84%E5%80%BCsin%CE%B1cos%CE%B11%2Fsin%CE%B1%2B1%2Fcos%CE%B1sin%E5%9B%9B%E6%AC%A1%E6%96%B9%CE%B1%2Bcos%E5%9B%9B%E6%AC%A1%E6%96%B9%CE%B1sin%26%23179%3B%CE%B1%2Bcos%26%23179%3B%CE%B1)
已知sinα,cosα是方程2x²-x+a=0的两个根,求下列各值sinαcosα1/sinα+1/cosαsin四次方α+cos四次方αsin³α+cos³α
已知sinα,cosα是方程2x²-x+a=0的两个根,求下列各值
sinαcosα
1/sinα+1/cosα
sin四次方α+cos四次方α
sin³α+cos³α
已知sinα,cosα是方程2x²-x+a=0的两个根,求下列各值sinαcosα1/sinα+1/cosαsin四次方α+cos四次方αsin³α+cos³α
因为 sina,cosa是方程2x²-x + a = 0的两个根
由韦达定理可知:
sina + cosa = 1/2 sinacosa = a/2
所以 sinacosa = a/2
1/sina + 1/cosa = (sina + cosa)/sinacosa = (1/2)/(a/2) = 1/a
sin^4 a +cos^4 a = (sin²a + cos²a)²-2sin²acos²a
= 1²-2(sinacosa)²
= 1 -2(a/2)²
= 1 - a²/2
sin³a+cos³a = (sina + cosa)(sin²a -sinacosa + cos²a)
= (1/2)(1-a/2)
= 1/2 - a/4