1
$\begingroup$

The problem is:

$\textrm{From the following equation}$

$$10\cos{2\omega}-13\cos{3\omega}+2\sin{\frac{5\omega}{2}}\sin{\frac{\omega}{2}}=0$$

$\textrm{find the value of:}$

$$\sec{\omega}+\sec{3\omega}$$ $\textrm{Consider:}$ $$\omega\neq (2K+1)\frac{\pi}{6}\;k\in \mathbb{Z}$$

I figured out some familiar expression in the third term of the equation and by applying prosthaphaeresis formula in the third term of the equation I got to this,

$10\cos{2\omega}-13\cos{3\omega}-(\cos3\omega-cos2\omega)=0$

$11\cos{2\omega}-14\cos{3\omega}=0$,

However there are second and double angles in the latter equation, therefore I decided to transform into their power equivalents shown below,

$11(\cos^{2}\omega-\sin^{2}\omega)-14(4\cos^{3}\omega-3cos\omega)=0$,

By using pitagorean identity in the earlier equation then I reached to this expression:

$11(\cos^{2}\omega-(1-\cos^2\omega))-14(4\cos^{3}\omega-3cos\omega)=0$,

$11(2\cos^{2}\omega-1)-14(4\cos^{3}\omega-3cos\omega)=0$,

$22\cos^{2}\omega-22-56\cos^{3}\omega+42\cos\omega=0$

to which is transformed into a cubic equation as follows (note I multiplied by $-1$):

$56\cos^{3}\omega-22\cos^{2}\omega-42\cos\omega+22=0$

I am not sure if my procedure is correct, moreover to solve a cubic equation entitles a problem since I don't know how to find the angle from there. How can I get to the answer?

$\endgroup$
1
$\begingroup$

Consider the equation $$11(2\cos^{2}\omega-1)-14(4\cos^{3}\omega-3\cos\omega)=0$$ and just multiply it by $2$: $$11(4\cos^{2}\omega-2)-28(4\cos^{3}\omega-3\cos\omega)=0$$ From this equation and the assumption that $\omega \neq (2k+1)\frac{\pi}{6}$, we conclude that $$\frac{4\cos^{2}\omega-2}{4\cos^{3}\omega-3\cos\omega}=\frac{28}{11}$$ But it is easy to check $ \frac{4\cos^{2}\omega-2}{4\cos^{3}\omega-3\cos\omega} = \sec \omega + \sec 3\omega $.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thanks for that, now by comparing with your reasoning I figured out that I was missing. It makes me glad to know I was not that far off from the answer. $\endgroup$ – Chris Steinbeck Bell Nov 4 '17 at 0:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.