Question

已知0＜a+b＜π/2 ,-π/2＜a-b＜π/3

已知0＜a+b＜π/2 ,-π/2＜a-b＜π/3 求2a和3a-（b/3）的取值范围

Answer 1

0＜a+b＜π/2 ,-π/2＜a-b＜π/3
两式相加得-π/2＜2a＜5π/6.

设3a-（b/3）=x(a+b)+y(a-b).
比较系数得：x+y=3,x-y=-1/3,
解得x=4/3,y=5/3.
3a-（b/3）=4/3 (a+b)+5/3 (a-b).
0＜a+b＜π/2 ,-π/2＜a-b＜π/3
则0＜4/3 (a+b)＜2π/3 ,-5π/6＜5/3 (a-b)＜5π/9，
两式相加得-5π/6＜4/3 (a+b)+5/3 (a-b) ＜11π/9
即-5π/6＜3a-（b/3）＜11π/9。