[15财富]高一数学 解三角形说明：一题5分,只答小题看情况给~1.三角形ABC中,2sin^2(A+B)/2+cos2

解:
1
2sin^2(A+B)/2+cos2c=2cos^2c/2+cos2c=cosc+1+cos2c=cosc+1+2cos^2c-1=cosc+2cos^2c=1
cosc=1/2或cosc=-1(不合题意,舍去),
所似C=60
余弦定理:a^2+b^2-2abcosC=3
5-2ab*1/2=3
ab=2
a=2/b
代入a2+b2=5
得:(1/b)^2+b^2=5
b=2,a=1或a=2或b=1
2
cosA=1/4
参考1:
sin^2(B+C)/2+cos2A=cosA+2cos^2A=1/4+1/8=3/8
余弦定理:c^2+b^2-2cbcosA=16
(b+c)^2=c^2+b^2+2bc=36
36-2bc-2bc*1/4=16
bc=40/3
联立:b+c=6,得:(自已算),
3:
A+C=180-B=2B,B=60
tg(A+C)=(tgA+tgC)/(1-tgAtgC)
-tg60=(tgA+tgC)/(1-2-根号3)
tgA+tgC=3+根号3
联立tgAtgC=2+根号3
tgA=1,tgC=2+根号3,或tgA=1,tgC=2+根号3,
因为C>A
所以,tgA=1 A=45
C=180-60-45=75
c=ctg45*4根号3+ctg60*4根号3=4根号3+3/4根号3