數學 M2 三角學 (二)?

2017-04-22 9:20 pm
求 Q.14,15 詳解

回答 (1)

2017-04-23 7:30 am
14
(a)
Cos^2 (2θ)
=(2Cos^2 θ-1)^2
=4Cos^4 θ-4Cos^2 θ+1
(b)
y=4Cos^2 (2θ)/[Cos^2 (2θ)+1]
=4/[1+Sec^2 (2θ)]
(c)
1+Sec^2 (2θ)>=1
1/[1+Sec^2 (2θ)<=1
4/[1+Sec^2 (2θ)<=4
y<=4
15
(a)
Sinθ
=2Sin(θ/2)Cos(θ/2)
=2Sin(θ/2)Cos(θ/2)/[Cos^2 (θ/2)+Sin^2 (θ/2)]
=2Tan(θ/2)/[1+Tan^2 (θ/2)]
=2t/(1+t^2)
(b)
Cosθ
=Cos^2 (θ/2)-Sin^2 (θ/2)
=[Cos^2 (θ/2)-Sin^2(θ/2)’/[Cos^2 (θ/2)+Sin^2 (θ/2)]
=[1-Tan^2 (θ/2)]/[1+Tan^2 (θ/2)]
=(1-t^2)/(1+t^2)
√2Sinθ+Cosθ=1
2√2t/(1+t^2)+(1-t^2)/(1+t^2)=1
2√2t+1-t^2=1+t^2
2t^2-2√2t=0
t=0 or t=√2
when t=0
Tan(θ/2)=0
θ=0
when t=√2
Tan(θ/2)=√2
θ/2=ArcTan√2
θ=2ArcTan√2


收錄日期: 2021-04-30 22:01:44
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