Question

在△ABC中，角A、B、C的对边分别为a、b、c，2sin^2A+C/2+cos2B=1，（1）若b=根号下13，a=3，求c=？

（2）设t=sinAsinc，当t取最大值时求A的值

Answer 1

你好！

2sin²[(A+C)/2] + cos2B = 1
1 - cos(A+C) + cos2B = 1
cosB + cos2B = 0
cosB + 2cos²B - 1 = 0
解得 cosB = -1 （舍去）或 cosB = 1/2
∴B = π/3

b² = a²+c² - 2accosB
13 = 9+c² - 3c
解得 c = 4

t = sinAsinC = sinA sin(2π/3 - A)
= sinA ( √3/2 cosA + 1/2 sinA)
= √3/4 sin2A + 1/4 ( 1 - cos2A)
= 1/2 sin(2A - π/6 ) + 1/4
0 < A < 2π/3
- π/6 < 2A - π/6 < 7π/6
当 2A - π/6 = π/2 即 A = π/3 时，t取最大值 3/4

【答案】
（1）c=4
（2）A = π/3