Chủ đề này có 2 trả lời

[luongkylinh]

Chứng minh rằng : 
$cos(\frac{2\pi }{2013})+cos(\frac{4\pi }{2013})+...+cos(\frac{2010\pi }{2013})+cos(\frac{2012\pi }{2013})=\frac{-1}{2}$

[Forgive Yourself]

Chứng minh rằng : 
$cos(\frac{2\pi }{2013})+cos(\frac{4\pi }{2013})+...+cos(\frac{2010\pi }{2013})+cos(\frac{2012\pi }{2013})=\frac{-1}{2}$

 

Ta có:

 

$2\sin \frac{\pi}{2013}.S=\sin\frac{3\pi}{2013}-\sin\frac{\pi}{2013}+\sin\frac{5\pi}{2013}-\sin\frac{3\pi}{2013}+...+\sin\pi-\sin\frac{2011\pi}{2013}=-\sin\frac{\pi}{2013}\\ \Leftrightarrow S=\frac{-1}{2}$

Bài viết đã được chỉnh sửa nội dung bởi Forgive Yourself: 07-01-2015 - 17:31

[Forgive Yourself]

Bài tổng quát:

$\forall k\epsilon N^*: 2\cos\frac{a}{2}\cos ka=\cos\frac{(2k-1)a}{2}+\cos\frac{(2k+1)a}{2}$
 

$\Rightarrow2\cos\frac{a}{2}\sum_{k=1}^{n}{(-1)^{k+1}\cos ka}=\sum_{k=1}^{n} {(-1)^{k+1}\cos\frac{(2k-1)a}{2}+(-1)^{k+1}\cos\frac{(2k+1)a}{2}}$
$=\cos\frac{a}{2}+(-1)^{n+1}\cos\frac{(2n+1)a}{2}$
$\Rightarrow\sum_{k=1}^{n}{(-1)^{k+1}\cos ka}=\frac{1}{2}+\frac{(-1)^{k+1}}{2}.\frac{\cos\frac{(2n+1)a}{2}}{cos\frac{a}{2}}$