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

来源:学生作业帮助网 编辑:作业帮 时间:2024/12/02 12:44:55
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
xSJPPs&YsWݘ"b!fe(EAXQZ+ꙻٔIw]]|9oj>͚ZG72n[Z>421{ysN8~g,rTed Ј4;d?JYͽ["djY*(W< , EVMݵwoyvf俬PEr>Qd+$Q@ DZЀ 8Gěan`-A4=>ש"S9eˊg)fHiqop%!R#@O2?"D?=*dPdԱM,Ss`>D`t " t4h:!pP!?11, (? }8cb|,-y$U.\7d

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
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

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
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