题目

在三角形ABC中,内角ABC对边为abc,满足√2asin(B+派/4)＝c
1,求角A.2,若为锐角三角形,求sinBsinC取值范围

提问时间：2021-01-03

答案

1
由题意,根据正弦定理,将原式中的a换成sinA,c换成sinC,得
√2sinA·sin(B+派/4)＝sinC ①
→sinA·(cosB+sinB)=sin(A+B)
→sinA·cosB+sinA·sinB=sinA·cosB+cosA·sinB
→sinA·sinB=cosA·sinB
→sinA=cosA
→tanA=1,
A=派/4.
2
sinBsinC=(-1/2)[cos(B+C)-cos(B-C)]
=(-1/2)[cos(π-A)-cos(B-C)]
=(-1/2)[cos(3π/4)-cos(B-C)]
=(-1/2)[√2/2 - cos(B-C)]
= -√2/4 + (1/2)cos(B-C)]
当B或C接近π/2时,C或B接近π/4.则|B-C|接近π/4.cos(B-C)接近√2/2.
当B=C时,cos(B-C)=1
而余弦函数在此区间内单调,
所以可得sinBsinC取值范围为
(0,1-√2/4]