# Thread:

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
Attached Thumbnails

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 $\cos(A+B) + \cos(A-B) = 2\cos A\cos B$ 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

4. Hello this is the solution :

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

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

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

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

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

Grandad

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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 23, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9443332552909851, "perplexity_flag": "middle"}