Question

在△ABC中，BC=a,AC=b,AB=c,a=√5,b=3,sinC=2sinA. （1）求AB

在△ABC中，BC=a,AC=b,AB=c,a=√5,b=3,sinC=2sinA. （1）求AB的值； （2）求sin(2A－π/4)的值

Answer 1

解：①由正弦定理，得 a/sinA=c/sinC ∴ c=a*sinC/sinA=2a=2√5
②由余弦定理贺和，得 cosA=(b²+c²-a²)/2bc=2√5/5,sinA=√(1-cos²A)=√5/5
sin(2A-π/4)=sin2A*cosπ/4-cos2A*sinπ/4
=2sinA*cosA*√2/2-(cos²禅早盯A-sin²A)*√2/2
=√睁余2/10

Answer 2

（1）∵ a = √局毁5 , b = 3 , sinC = 2sinA
∴ a ：c = sinA ：sinC
√5 ：c = sinA ：2sinA
√5 ：c = 1 ：2

c = 2 √明芦5
∴AB = 2 √5
（2）∵激腊带sinA = a/c =1/2
∴A = π/6
∴sin(2A－π/4)
= sin(π/3－π/4)
= sinπ/3·cosπ/4-cosπ/3·sinπ/4
= √3/2 × √2/2 - 1/2 × √2/2
= √6/4 - √2/4

Answer 3

①2√5
②√2/10