已知-∏/2≤α<β≤∏/2,求α+β/2 α-β/2的取值范圍?
熱心網友
(1)-π/2≤α<π/2 -π/2<β≤π/2所以 -π<α+β<π -π/2<(α+β)/2<π/2(2)-π/2≤α<π/2 -π/2≤-β<π/2所以-π≤α-β<π -π/2≤(α-β)/2<π/2注:最主要考慮是否能取等號.
已知-∏/2≤α<β≤∏/2,求α+β/2 α-β/2的取值范圍?
(1)-π/2≤α<π/2 -π/2<β≤π/2所以 -π<α+β<π -π/2<(α+β)/2<π/2(2)-π/2≤α<π/2 -π/2≤-β<π/2所以-π≤α-β<π -π/2≤(α-β)/2<π/2注:最主要考慮是否能取等號.