Solve for A: cos 9A = (cos 15A)(cos 8A)
Hello pacmanI 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!
Grandad
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}
$
$\displaystyle \begin{array}{l}
subst: \\
\cos \left( a \right)\left( {1 - \cos \left( {8a} \right)} \right) = 0 \\
\cos (a) = 0:a = \frac{\pi }{2} + \pi k...k \in Z \\
\cos \left( {8a} \right) = 1:a = \frac{{\pi k}}{4}...k \in Z \\
\end{array}$