1. ## trigonometric equation

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

2. 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

3. 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 $\displaystyle \cos(A+B) + \cos(A-B) = 2\cos A\cos B$ on the RHS and say:

$\displaystyle 2\cos9A = 2\cos15A\cos8A$

$\displaystyle = \cos23A + \cos7A$

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

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

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

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

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

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

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

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

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

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

4. Hello this is the solution :

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

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

5. Hello dhiab
Originally Posted by dhiab
Hello this is the solution :

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

$\displaystyle \begin{array}{l} subst: \\ \cos \left( a \right)\left( {1 - \cos \left( {8a} \right)} \right) = 0 \\ \cos (a) = 0:a = \frac{\pi }{2} + \pi k...k \in Z \\ \cos \left( {8a} \right) = 1:a = \frac{{\pi k}}{4}...k \in Z \\ \end{array}$
Sorry, but this doesn't make sense. You can't simply replace $\displaystyle \cos9A$ and $\displaystyle \cos15A$ by $\displaystyle \cos A$.

6. 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