Question

求数学三角函数的二倍角公式…… 急用！！！

Answer 1

二倍角公式：　　sin(2α)=2sinα·cosα=2tan(α)/[1+tan^2(α)]
　　cos(2α)=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)=[1-tan^2(α)]/[1+tan^2(α)]
　　tan(2α)=2tanα/[1-tan^2(α)]
由二倍角引申出来的：半角公式：　　
sin^2(α/2)=(1-cosα)/2
　　cos^2(α/2)=(1+cosα)/2
　　tan^2(α/2)=(1-cosα)/(1+cosα)
　　tan(α/2)=sinα/(1+cosα)=(1-cosα)/sinα

Answer 2

sin2x=2sinxcos
cos2x=2cos^2x-1=1-2sin^2x
tan2x=2tanx/(1-tan^2x)

Answer 3

sin(2a)=sin(a+a) =sinacosa+cosasina=2sinacosa
Cos(2a)=cos(a+a)=cosacosa-sinasina=cos²a-sin²a.
Sin(3a)=sin(2a+a) =sin2acosa+cos2asina=2sinacosa cosa+( cos²a-sin²a) sina=2sina(1- sin²a) +(1- 2sin²a) sina=3sina-4sin³a
Cos(3a)= cos(2a+a) = cos2acosa-sin2asina=( cos²a-sin²a) cosa-2sinacosasina=( 2cos²a-1) cosa-2cosa(1- cos²a)=4cos³a-3coa.

Answer 4

　sin2α = 2cosαsinα
1.cos2α = 2cos^2 α- 1 　　2.cos2α = 1 − 2sin^2 α 　　3.cos2α = cos^2 α − sin^2 α
　tan2α = 2tanα/[1 - (tan^2α)] 　　tan(1/2*α)=(sin α)/(1+cos α)=(1-cos α)/sin α