已知a,b,c为正角，(sina)^2+(sinb)^2+(sinc)^2=1,求证：a+b+c>90度

如果a+b+c<=90°
则a,b,c均为锐角
则cos(a-b)>cos(a+b)
1=(sina)^2+(sinb)^2+(sinc)^2
=1-(cosa)^2+(sinb)^2+(sinc)^2
=1-cos(a+b)cos(a-b)+(sinc)^2
<1-[cos(a+b)]^2+(sinc)^2
即[cos(a+b)]^2<(sinc)^2
0<a+b<90 0<c<90
cos(a+b)<sinc
而：a+b+c<=90 0<c<=90-a-b<90
sinc<=sin(90-a-b)=cos(a+b)
得出矛盾！因此假设不成立！

先反过来分析：
假设a+b+c≤90°，易知a+b+c>0°
于是，0°<a+b≤90°-c<90°
0<sin(a+b)≤sin(90°-c)=cosc
∴(sin(a+b))^2+(sinc)^2≤(cosc)^2+(sinc)^2=1
若还能证明：(sina)^2+(sinb)^2<(sin(a+b))^2，则找到了矛盾
(sin(a+b))^2=(sinacosb+cosasinb)^2=(sinacosb)^2+2sinacosbcosasinb+(cosasinb)^2
(sina)^2+(sinb)^2=(sina)^2((sinb)^2+(cosb)^2)+(sinb)^2((sina)^2+(cosa)^2)=(sinacosb)^2+(sinbcosa)^2+2(sinasinb)^2
只需证：sinacosbcosasinb>(sinasinb)^2，即cosacosb>sinasinb
实际上，0°<a≤90°-b-c<90°-b<90°
∴0<sina<sin(90°-b)=cosb
同理：0<sinb<cosa
于是得到cosacosb>sinasinb