数学 三角関数

2017/5-2012/1 Yuji.W

☆三角関数の微分☆

 _〔物理定数

◇ ベクトル <A> 単位ベクトル <-u> 内積 * 外積 # 微分 y;x 時間微分 x'
 定積分 ${f(x)*dx}[x:a~b] 10^x=Ten(x) exp(i*x)=expi(x)

◇三角関数の微分◇

● lim[h->0]{sin(h)/h}=1

 sin(x+h)=sin(x)*cos(h)+cos(x)*sin(h)

 cos(a)=1-2*sin(a/2)^2

◎ sin(x);x を求めよう。

■ [cos(h)-1]/h=-2*sin(h/2)^2/h=-(1/2)*[sin(h/2)/(h/2)]^2*h だから、

 lim[h->0]{[cos(h)-1]/h}
=-(1/2)*1*0=0

 sin(x+h)-sin(x)=sin(x)*[cos(h)-1]+cos(x)*sin(h)

 sin(x);x
=lim[h->0]{[sin(x+h)-sin(x)]/h}
=0+cos(x)*1
=cos(x)
 

◎ cos(x);x を求めよう。

● cos(a+b)=Ca*Cb-Sa*Sb

● lim[h->0]{[cos(h)-1]/h}=0

■ cos(x+h)-cos(x)
=cos(x)*[cos(h)-1]-sin(x)*sin(h)

 cos(x);x
=lim[h->0]{[cos(x+h)-cos(x)]/h}
=0-sin(x)*1
=-sin(x)
 

■ tan(x);x=1/cos(x)^2 ・{覚えてなかった!2013/10}

※${(1/cos(x)^2)*dx}=tan(x)

{証明}tan(x);x
=(sin(x)/cos(x));x
=[(sin(x);x)*cos(x)-sin(x)*(cos(x);x)]/cos(x)^2
=(cos(x)^2+sin(x)^2)/cos(x)^2
=1/cos(x)^2

■ sec(x);x=(1/cos(x));x=sin(x)/cos(x)^2
 cosec(x);x=(1/sin(x));x=-cos(x)/sin(x)^2
 cot(x);x=(cos(x)/sin(x));x=-1/sin(x)^2

inserted by FC2 system