# trigonometric equation

• Sep 8th 2009, 11:30 PM
pacman
trigonometric equation
Solve for A: cos 9A = (cos 15A)(cos 8A)
• Sep 8th 2009, 11:58 PM
pacman
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
• Sep 9th 2009, 01:45 AM
Hello pacman
Quote:

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!

• Sep 9th 2009, 04:26 AM
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}
$

$\begin{array}{l}
subst: \\
\cos \left( a \right)\left( {1 - \cos \left( {8a} \right)} \right) = 0 \\
\cos (a) = 0:a = \frac{\pi }{2} + \pi k...k \in Z \\
\cos \left( {8a} \right) = 1:a = \frac{{\pi k}}{4}...k \in Z \\
\end{array}$
• Sep 9th 2009, 04:32 AM
Hello dhiab
Quote:

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 $\cos A$.

• Sep 9th 2009, 05:41 AM
pacman
dhiab, this puzzled me,

http://www.mathhelpforum.com/math-he...828d72bc-1.gif

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

But THANKS though, it cracks my mind