Question

设△ABC的内角A、B、C的对边长分别为a、b、c，且3b 2 +3c 2 -3a 2 =4 2 bc．（Ⅰ）求sinA

设△ABC的内角A、B、C的对边长分别为a、b、c，且3b 2 +3c 2 -3a 2 =4 2 bc．（Ⅰ）求sinA的值；（Ⅱ）求 2sin(A+ π 4 )sin(B+C+ π 4 ) 1-cos2A 的值．

Answer 1

（Ⅰ）由余弦定理得 cosA= b 2 + c 2 - a 2 2bc = 2 2 3 又 0＜A＜π，故sinA= 1- cos 2 A = 1 3 （Ⅱ）原式= 2sin(A+ π 4 )sin(π-A+ π 4 ) 1-cos2A = 2sin(A+ π 4 )sin(A- π 4 ) 2 sin 2 A = 2( 2 2 sinA+ 2 2 cosA)( 2 2 sinA- 2 2 cosA) 2 sin 2 A = sin 2 A- cos 2 A 2 sin 2 A = - 7 2 ．