Re: [問題] 三角函數
※ 引述《skyluck (蔥花)》之銘言：
: sin(pi/7)xsin(2*pi/7)xsin(3*pi/7)=?
: 和差化積
: 兩倍角公式
: 三倍角公式
: 一直算不出來
: 請板上各位大大幫忙了 Orz
Given s=sin, c=cos, and t=pi/7,
Find L=s(t)*s(2t)*s(3t)=?
Solution:
From 7*t=pi, we can develope as follows:
4*t=pi-3*t
s(4t)=s(pi-3t)
2*s(2t)*c(2t)=s(3t)
4*s(t)*c(t)*[2*c(t)^2-1]=s(t)*[3-4*s(t)^2].
Let u=s(t) and v=c(t), we can get
4*v*(2*v^2-1)=3-4*(1-v^2)
8*v^3-4*v=4-4*v^2-3
8*v^3-4*v^2+4*v+1=0.
It can be solved by Newton's method:
cos(t)=0.901, and
sin(t)=sr(1-v^2)=0.434,
where sr=square root.
Then sin(2t)=2*u*v=2*0.434*0.901=0.782,
and sin(3t)=u*(3-4*u^2)=0.434*(3-4*0.434^2)=0.975.
So it can be written as L=s(t)*s(2t)*s(3t)=0.331.
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