1
$\begingroup$

Presently I am faced with the following question:

By showing that $$\cos5\theta = \cos\theta(16\cos^4\theta - 20\cos^2\theta + 5)$$ and then solving the equation $\cos5\theta = 0$, deduce that

$$\cos^2\left(\frac{\pi}{10}\right) = \frac{5+\sqrt{5}}{8}$$

and hence find the exact values of $\cos^2\left(\frac{3\pi}{10}\right), \cos^2\left(\frac{7\pi}{10}\right), \cos^2\left(\frac{9\pi}{10}\right)$.

I have proved the first part using de Movire's theorem and after solving $\cos5\theta = 0$ reached the following equation:

$$\cos^2 \left(\frac{\pi}{10}\right) = \frac{5\pm\sqrt{5}}{8}$$

Where $\frac{\pi}{10}$ can be replaced with $\frac{3\pi}{10}$, $\frac{7\pi}{10}$, etc. as they are other possible solutions when finding $\theta$ from $\cos5\theta = 0$.

However I am unsure how I am supposed to know whether or not to take the positive or negative root dependent on the angle (i.e. for $\frac{\pi}{10}$, $\frac{7\pi}{10}$, $\frac{3\pi}{10}$, $\frac{9\pi}{10}$), in order to get the exact value.

$\endgroup$
1
$\begingroup$

Since $\cos^2 \theta$ is strictly decreasing on $\left[0, \frac{\pi}{2}\right]$, we must have that $$\cos^2 \left(\frac{\pi}{10}\right) > \cos^2 \left(\frac{3 \pi}{10}\right),$$ and so $$\cos^2 \left(\frac{\pi}{10}\right) = \frac{5 + \sqrt{5}}{8} \qquad \text{and} \qquad \cos^2 \left(\frac{3\pi}{10}\right) = \frac{5 - \sqrt{5}}{8}.$$ The remaining values can be determined by a similar argument, or by using these values together with the easy identity $\cos^2 \theta = \cos^2 (\pi - \theta)$.

$\endgroup$
  • $\begingroup$ That makes perfect sense, thought it may have been something obvious I was looking over. Thanks! $\endgroup$ – Joshua Nov 16 '14 at 15:45
  • $\begingroup$ You're welcome, I'm glad you found it useful! $\endgroup$ – Travis Nov 16 '14 at 16:27

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.