cos a =1 - 2(sina/2)^2; cos b =1 - 2(sin b/2)^2.
cosa + cosb - 2√2sina/2*sinb/2 =1 - 2(sina/2)^2 +1 - 2(sin b/2)^2 -2√2sina/2*sinb/2 =1
即 1 =2(sin a/2)^2 + 2(sin b/2)^2 + 2√2sina/2 * sinb/2
由积化和差公式：sinαsinβ=-[cos(α+β)-cos(α-β)]/2
代入得,cosa+cosb+2√2*[cos(sin a/2+sin b/2)-cos(sin a/2-sin b/2)]/2 =1
cosa + cosb -√2*[cos(sin a/2+sin b/2)-cos(sin a/2-sin b/2)]=1……
这一题好难啊.假设a + b =90°,代入可以得到上式成立.
所以三角形abc为直角三角形.希望有更好的解答~