Question

在△abc中b=3 c=1 A=2B①求a②求sin(A+π/4)

Answer 1

正弦定理：
3/sinb = a/sin2b = 1/sin3b = R
可得：
3 = r sinb
a = r sin2b
1 = r sin3b
第一式和第二式相除,运用倍角公式,可得 a/3 = 2cosb
第二式和第三式相除,a = sin2b / sin3b,运用倍角公式以及三倍角公式,a = 2cosb / (4(cosb)^2 - 1),再带入a/3 = 2cosb,所以a = 2√3
所以cosb = √3/3,sinb = √6/3
sina = sin2b = 2√2/3, cosa = cos2b = -1/3
sin(A+π/4) = 2√2/3 * √2/2 - 1/3 * √2/2 = (4 - √2) / 6