1已知2sinθ=1+2cosθ且θ為銳角則cos2θ/sin(θ-45°) 2已知270°<θ<360°且cos2θ及sinθ為方程式9x²-7x+k=0則tanθ/2 3ΔABC中,角A角B角C的對邊之邊長分別為a、b、c,已知sinC+cosC=1-sinC/2求sinC?

2018-08-28 9:37 pm

回答 (1)

2018-08-29 6:09 pm
1.已知2Sinθ=1+2Cosθ,且θ為銳角,則Cos2θ/Sin(θ-45°)?
Sol
2Sinθ=1+2Cosθ
4Sin^2 θ=1+4Cosθ+4Cos^2 θ
4-4Cos^2 θ=1+4Cosθ+4Cos^2 θ
8Cos^2 θ+4Cosθ-3=0
(2Cosθ+3)(2Cosθ-1)=0
Cosθ=1/2
θ為銳角
θ=60°
Cos2θ=Cos120 °=-1/2
Sin(θ-45°)=Sin15°=(√6-√2)/4
Cos2θ/Sin(θ-45°)
=(-1/2)/ [(√6-√2)/4]
=-2/(√6-√2)
=-2(√6+√2)/4
=-(√6+√2)/2
2.已知270°<θ<360°且Cos2θ及Sinθ為方程式9x^2-7x+k=0的兩根,則Tan(θ/2)?
Sol
Cos2θ+Sinθ=7/9
18Cos2θ+18Sinθ=14
18(1-2Sin^2 θ)+18Sinθ=14
-36Sin^2 θ+18Sinθ+4=0
36Sin^2 θ-18Sinθ-4=0
18Sin^2 θ-9Sinθ-2=0
(3Sinθ-2)(6Sinθ+1)+0
270°<θ<360°
Sinθ=2/3(不合) or Sinθ=-1/6
135°<θ/2<180°
A=Tan(θ/2)<0
Sinθ=2A/(1+A^2)=-1/6
A^2+1=-12A
A^2+12A+1=0
A=(-12+/-√140)/2=-6+/-√35
3ΔABC中,角A角B角C的對邊之邊長分別為a、b、c,已知SinC+CosC=1-Sin(C/2),求SinC?
Sol
P=C/2
SinC+CosC=1-Sin(C/2)
Sin(2P)+Cos(2P)=1-SinP
2SinPCosP+1-Sin^2 P=1-SinP
2SinPCosP-Sin^2 P+SinP=0
SinP<>0
2CosP-SinP+1=0
2CosP=SinP-1
4Cos^2 P=Sin^2 P-2SinP+1
4-4Sin^2 P=Sin^2 P-2SinP+1
5Sin^2 P-2SinP-3=0
(5SinP+3)(SinP-1)=0
SinP=-3/5 or SinP=1(0<P<π/2不合)
CosP=4/5
SinC=2sinPCosP=-24/25


收錄日期: 2021-04-30 22:41:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180828133702AAtOkqV

檢視 Wayback Machine 備份