- 小数学题`sinA+sinB=1/2CosA+cosB范围
- sinA+sin B=1/2
osA+cosB范围
- sinA + sinB = 1/2 <1>
设cosA + cosB = a <2>
<1>^2 + <2>^2得
2(sinAsinB + cosAcosB) = a^2 - 7/4
因为2sinAsinB = -cos(A + B) + cos(A - B)
2cosAcosB = cos(A + B) + cos(A - B)
即cos(A - B) = (a^2 - 7/4)/2
由于-1≤cos(A - B)≤1
所以-1≤(a^2 - 7/4)/2≤1
由此求得
a≥-√15/2 或 a≤√15/2