M_SHIRAISHIさんの<800c7853.0406110532.3b5138a8@posting.google.com>から
>Yoshitaka Ikeda <ikeda@4bn.ne.jp> wrote in message 
>news:<ca9564$eai$1@caraway.media.kyoto-u.ac.jp>...
>> ronさんの<c9sjf1$2fh$1@inews.gazeta.pl>から
>> >how can I simplify
>> >
>> >sin(x) + sin(2x) + sin(3x)
>> >cos(x) + cos(2x) + cos(3x)
>> >
>> >greetings
>> 
>> sin(x) can be written sin(2x-x)
>> sin(3x) can be written sin(2x+x)
>> So, 
>> sin(x)=sin(2x)cos(x)-cos(2x)sin(x)
>> and 
>> sin(3x)=sin(2x)cos(x)+cos(2x)sin(x)
>> 
>> So,
>> sin(x)+sin(2x)+sin(3x)=sin(2x)(2cos(x)+1)
>> 
>> 
>> In a similar way
>> cos(x)=cos(2x-x)=cos(2x)cos(x)+sin(2x)sin(x)
>> cos(3x)=cos(2x+x)=cos(2x)cos(x)-sin(2x)sin(x)
>> 
>> cos(x)+cos(2x)+cos(3x)=cos(2x)(2cos(x)+1)
>
>
>However, "sin(x) + sin(2x) + sin(3x)"looks more simple 
>than"sin(2x)(2cos(x)+1)"
>isn't it ?

First, I thought the problem is to simplify 
(sin(x)+sin(2x)+sin(3x))/(cos(x)+cos(2x)+cos(3x)).
So that, which I wrote was not a finally answer.
This problem's answer is sin(2x)/cos(2x).

After I wrote it, I thought that if the problem is to simplify "both" 
expressions, there will be a problem which is simple 
"sin(x)+sin(2x)+sin(3x)" or "sin(2x)(2sin(x)+1)".
But I thought "sin(2x)(2sin(x)+1)" is more simple.
Because, there's only two sin function.
In addition, if there's a equation "sin(x)+sin(2x)+sin(3x)=0",
we can get a solution that "sin(2x)=0" or "sin(x)=0.5" which is from 
"sin(2x)(2sin(x)+1)=0".


-- 
Yoshitaka Ikeda mailto:ikeda@4bn.ne.jp