trigonometric equation

**pacman** · September 8th 2009, 10:30 PM

Solve for A: cos 9A = (cos 15A)(cos 8A)

**pacman** · September 8th 2009, 10:58 PM

i have tried to use a math program, and the graph looks like this, the red dot are the roots. I think there so many of them

**Grandad** · September 9th 2009, 12:45 AM

Hello pacman

Originally Posted by pacman

Solve for A: cos 9A = (cos 15A)(cos 8A)

I can't see how to find the complete solution yet, but the most likely approach is probably to use $\cos(A+B) + \cos(A-B) = 2\cos A\cos B$ on the RHS and say:

$2\cos9A = 2\cos15A\cos8A$

$= \cos23A + \cos7A$

$\Rightarrow \cos9A - \cos7A= \cos23A -\cos9A$

Now use $\cos A - \cos B = -2\sin\tfrac12(A+B)\sin\tfrac12(A-B)$ on each side:

$-2\sin8A\sin A=-2\sin16A\sin7A$

$\sin8A\sin A= 2\sin8A\cos8A\sin7A$

$\sin8A = 0$ or $\sin A = 2\cos8A\sin7A$

So that, at least, gives some solutions: $\sin8A=0\Rightarrow A = \frac{n\pi}{8}$

But what happens next with the remaining expression, I'm not sure. You could use $\sin7A = \sin A(7-56\sin^2A+112\sin^4A-64\sin^6A)$ and then remove the factor $\sin A$ from both sides to get:

$\sin A=0$ (which doesn't give any more answers that we haven't already got from $\sin8A=0$ ), or:

$1 = 2\cos8A(7-56\sin^2A+112\sin^4A-64\sin^6A)$

which you could then express simply in terms of $\cos A$ , but it would be pretty horrendous!

Grandad

**dhiab** · September 9th 2009, 03:26 AM

Hello this is the solution :

$\begin{array}{l} \cos \left( {9a} \right) - \cos \left( {15a} \right)\cos \left( {8a} \right) = 0 \\ but: \\ \cos \left( {9a} \right) = \cos \left( {8a + a} \right) = \cos (a) \\ \cos \left( {15a} \right) = \cos \left( {16a - a} \right) = \cos \left( { - a} \right) = \cos \left( a \right) \\ \end{array} $

$\begin{array}{l} subst: \\ \cos \left( a \right)\left( {1 - \cos \left( {8a} \right)} \right) = 0 \\ \cos (a) = 0:a = \frac{\pi }{2} + \pi k...k \in Z \\ \cos \left( {8a} \right) = 1:a = \frac{{\pi k}}{4}...k \in Z \\ \end{array}$

**Grandad** · September 9th 2009, 03:32 AM

Hello dhiab

Originally Posted by dhiab

Hello this is the solution :

$\begin{array}{l} \cos \left( {9a} \right) - \cos \left( {15a} \right)\cos \left( {8a} \right) = 0 \\ but: \\ \cos \left( {9a} \right) = \cos \left( {8a + a} \right) = \cos (a) \\ \cos \left( {15a} \right) = \cos \left( {16a - a} \right) = \cos \left( { - a} \right) = \cos \left( a \right) \\ \end{array} $

$\begin{array}{l} subst: \\ \cos \left( a \right)\left( {1 - \cos \left( {8a} \right)} \right) = 0 \\ \cos (a) = 0:a = \frac{\pi }{2} + \pi k...k \in Z \\ \cos \left( {8a} \right) = 1:a = \frac{{\pi k}}{4}...k \in Z \\ \end{array}$

Sorry, but this doesn't make sense. You can't simply replace $\cos9A$ and $\cos15A$ by $\cos A$ .

Grandad

**pacman** · September 9th 2009, 04:41 AM

dhiab, this puzzled me,

a is not 20 degrees or what, as GRANDAD have said your post sems puzzling.

But THANKS though, it cracks my mind

**pacman** · September 9th 2009, 06:34 AM

Mathematica gave this factorization, but i can not arrive to it, please help.

32 sin^2(A) sin(A-pi/4) sin(2A-pi/4) sin(A+pi/4) sin(2A+pi/4) cos(A) (2cos(2A)-1) (2cos(6A)+4 cos(8A)+4 cos(10A)+2 cos(12A)+1) = 0

**pacman** · September 9th 2009, 06:36 AM

a few more plot of the said equation,

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