若A属于(0,派),且SINA+COSA=1/3,求COS2A的值

1/3 = sinA + cosA = √2·[(sinA)/√2 + (cosA)/√2] = √2·sin(A + π/4) ,∴sin(A + π/4) = √2/6 ,∵A∈(0 ,π) ,∴A + π/4 ∈(π/4 ,5π/4) ,但sin(A + π/4) > 0 ,∴A + π/4 ∈(π/4 ,π) ,又∵√2/6 < √2/2 ,∴根据正弦函数的单调性可知：A + π/4 < π/4 或 A + π/4 > 3π/4 ,∵前一个结论与“A∈(0 ,π)”矛盾 ,∴A + π/4 > 3π/4 ,∴A > π/2 ,又∵A + π/4 ∈(π/4 ,π) ,∴A < 3π/4 ,∴A的范围是(π/2 ,3π/4) ,∴2A∈(π ,3π/2) ,在此区间内 ,cos2A < 0 .
对“1/3 = sinA + cosA ”两边平方：1/9 = 1 + 2sinAcosA = 1 + sin2A ,∴sin2A = -8/9 ,代入同角关系式并结合 cos2A < 0 得：cos2A = -(√17)/9 .