sina+cosa=1/5,求sina.cosa

设 x =cos a,y =sin a,
则 x +y =1/5,(1)
x^2 +y^2 =1.(2)
(1)^2 -(2) 得
2 xy = -24/25,
即 xy = -12/25.(3)
由(1)(3) 可知,x,y 是方程
z^2 -(1/5)z -12/25 =0.(4)
的两根.
解(4)得
z1 =4/5,z2 = -3/5.
所以 sin a =4/5,cos a = -3/5,
或 sin a = -3/5,sin a =4/5.
= = = = = = = = =
换元法,解方程.
同角三角函数问题,实质上是代数问题.