trigo-sum

dhiab · #1 $Old$ June 16th, 2009, 09:09 AM

Calculate this sum :

pickslides · #2 $Old$ June 17th, 2009, 03:47 PM

you need to evaluate $cos(\frac{\pi}{9})$

To do this you need to consider $cos(a\pm b) = cos(a)\times cos(b)\mp sin(a)\times sin(b)$

where $cos(\frac{\pi}{9}) = cos(a\pm b)$

only after finding this will you be able to get an exact value

mr fantastic · #3 $Old$ June 17th, 2009, 09:37 PM

Quote:

Originally Posted by pickslides $View Post$

you need to evaluate $cos(\frac{\pi}{9})$

To do this you need to consider $cos(a\pm b) = cos(a)\times cos(b)\mp sin(a)\times sin(b)$

where $cos(\frac{\pi}{9}) = cos(a\pm b)$

only after finding this will you be able to get an exact value

This technique will not work in this instance.

Edit: The off-topic discussion regarding the value of questions like this one can be found here: 18th century questions in the 21st century. Any further discussion of this nature should be continued at that thread.

dhiab · #4 $Old$ June 17th, 2009, 11:16 PM

help: use relation

Bruno J. · #5 $Old$ June 18th, 2009, 12:02 AM

Quote:

Originally Posted by dhiab $View Post$

help: use relation

Or you can try

$(a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2)=$

$(a_1 b_1 - a_2 b_2 - a_3 b_3 - a_4 b_4)^2 +(a_1 b_2 + a_2 b_1 + a_3 b_4 - a_4 b_3)^2 +$

$(a_1 b_3 - a_2 b_4 + a_3 b_1 + a_4 b_2)^2 +(a_1 b_4 + a_2 b_3 - a_3 b_2 + a_4 b_1)^2$

although I don't promise anything.

mr fantastic · #6 $Old$ June 18th, 2009, 12:21 AM

Quote:

Originally Posted by dhiab $View Post$

help: use relation

That thought had crossed my mind but it will then be necessary to solve the cubic equation $4 t^3 - 3t - \frac{1}{2} = 0$ and I think that will be no mean feat (the method of Cardano could be used I suppose).

There will be a clever trick but I don't have time right.

malaygoel · #7 $Old$ June 18th, 2009, 08:47 PM

Let us consider the equation
$cos3\theta=cos\frac{\pi}{3}$

there are infinite solutions of this equation
in the domain $[0,2\pi]$ , the solutions are

$\theta=\frac{\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9},\frac{11\pi}{9},\frac{13\pi}{9},\frac{17\pi}{9}.................(a)$

Quote:

Equation1:
$cos3\theta=cos\frac{\pi}{3}$
Equation 2:
$4cos^{3}x-3cosx=1/2$

The above two equations are identical. Every value of $\theta$ that satisfies equation 1 satisfies equation 2. If $\theta$ is a solution of equation 1, the $cos\theta$ is a root of the second equation.

For above solutions of $\theta$

$cos\theta=cos\frac{\pi}{9},cos\frac{5\pi}{9},cos\frac{7\pi}{9},cos\frac{11\pi}{9},cos\frac{13\pi}{9},cos\frac{17\pi}{9}$
the last three of the above six are just the repetition of the first three. So the roots of Equation 2 can be

$cos\frac{\pi}{9}(=cos\frac{17\pi}{9}),cos\frac{5\pi}{9}(=cos\frac{13\pi}{9}),cos\frac{7\pi}{9}(=cos\frac{11\pi}{9})$

Hence, we can say that $cos\frac{\pi}{9},cos\frac{5\pi}{9},cos\frac{7\pi}{9}$ are the three roots of equation 2.

Multipying equation 2 by $sec^{3}\theta$
we get

$sec^{3}\theta+6sec^{2}\theta-8=0$

roots are $sec\frac{\pi}{9},sec\frac{5\pi}{9},sec\frac{7\pi}{9}$

$sec\frac{\pi}{9}+sec\frac{5\pi}{9}+sec\frac{7\pi}{9}=-6$

$sec\frac{\pi}{9}.sec\frac{5\pi}{9}+sec\frac{7\pi}{9}.sec\frac{\pi}{9}+sec\frac{5\pi}{9}.sec\frac{7\pi}{9}=0$

the question of this thread is
$sec^{2}\frac{\pi}{9}+sec^{2}\frac{5\pi}{9}+sec^{2}\frac{7\pi}{9}+4$

$=(sec\frac{\pi}{9}+sec\frac{5\pi}{9}+sec\frac{7\pi}{9})^{2}-2(sec\frac{\pi}{9}.sec\frac{5\pi}{9}+sec\frac{7\pi}{9}.sec\frac{\pi}{9}+sec\frac{5\pi}{9}.sec\frac{7\pi}{9})+4$

$=(-6)^{2}+2(0)+4$

$=40$

dhiab · #8 $Old$ June 19th, 2009, 01:12 AM

Quote:

Originally Posted by mr fantastic $View Post$

This technique will not work in this instance.

Edit: The off-topic discussion regarding the value of questions like this one can be found here: 18th century questions in the 21st century. Any further discussion of this nature should be continued at that thread.

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