1
由題意,根據(jù)正弦定理,將原式中的a換成sinA,c換成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)]
當B或C接近π/2時,C或B接近π/4.則|B-C|接近π/4.cos(B-C)接近√2/2.
當B=C時,cos(B-C)=1
而余弦函數(shù)在此區(qū)間內(nèi)單調(diào),
所以可得sinBsinC取值范圍為
(0,1-√2/4]