trigo-sum

**dhiab** · Jun 16th 2009, 07:09 AM

Calculate this sum :

**pickslides** · Jun 17th 2009, 01:47 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** · Jun 17th 2009, 07:37 PM

Originally Posted by pickslides

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: http://www.mathhelpforum.com/math-he...t-century.html. Any further discussion of this nature should be continued at that thread.

**dhiab** · Jun 17th 2009, 09:16 PM

help: use relation

**Bruno J.** · Jun 17th 2009, 10:02 PM

Originally Posted by dhiab

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** · Jun 17th 2009, 10:21 PM

Originally Posted by dhiab

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** · Jun 18th 2009, 06:47 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

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

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 $\theta$

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

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

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

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

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

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** · Jun 18th 2009, 11:12 PM

Originally Posted by mr fantastic

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: http://www.mathhelpforum.com/math-he...t-century.html. Any further discussion of this nature should be continued at that thread.

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