设锐角三角形ABC的内角A,B,C的对边分别为a,b,c,a^2+c^2-b^2=根号3ac 求cosA+sinC的取值范围
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![设锐角三角形ABC的内角A,B,C的对边分别为a,b,c,a^2+c^2-b^2=根号3ac 求cosA+sinC的取值范围](/uploads/image/z/3715399-55-9.jpg?t=%E8%AE%BE%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%86%85%E8%A7%92A%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2Ca%5E2%2Bc%5E2-b%5E2%3D%E6%A0%B9%E5%8F%B73ac+%E6%B1%82cosA%2BsinC%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
设锐角三角形ABC的内角A,B,C的对边分别为a,b,c,a^2+c^2-b^2=根号3ac 求cosA+sinC的取值范围
设锐角三角形ABC的内角A,B,C的对边分别为a,b,c,a^2+c^2-b^2=根号3ac 求cosA+sinC的取值范围
设锐角三角形ABC的内角A,B,C的对边分别为a,b,c,a^2+c^2-b^2=根号3ac 求cosA+sinC的取值范围
a²+c²-b²=√3ac
由余弦定理得
cosC=(a²+c²-b²)/(2ac)=√3/2
C=π/6
A+B+C=π,A=π-C-B=π-π/6-B=5π/6-B
A>0,B>0,0
由b^2=a^2+c^2-2accosB
知cosB=√3/2 B=30°
A=180°-30°-C=150°-C
0cosA+sinC=cos(150°-C)+sinC
=sinC+sin(C-60°)
=sin(C-30°)
可见 1/2≤cosA+sinC≤1