4
$\begingroup$

Find the value of $\tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\tan^2\dfrac{4\pi}{16}+\tan^2\dfrac{5\pi}{16}+\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}.$

My attempts:

I converted the given series to a simpler form:

$\tan^2\dfrac{\pi}{16}+\cot^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\cot^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\cot^2\dfrac{3\pi}{16}+1.$

Then I found the following values because I already knew the values of $\sin22.5^{\circ}$ and $\cos22.5^{\circ}$:

$\cos^2(\frac{\pi}{16})= \dfrac{2+\sqrt{2+\sqrt2}}{4}$

$\sin^2(\frac{\pi}{16})= \dfrac{2-\sqrt{2+\sqrt2}}{4}$

$\sin^2(\frac{\pi}{8})= \dfrac{2-\sqrt2}{4}$

$\cos^2(\frac{\pi}{8})= \dfrac{2+\sqrt2}{4}$

However, at this stage I feel that my method of solving this problem is unnecessarily long and complicated. Could you guide me with a simpler approach to this question?

$\endgroup$
2
$\begingroup$

$$\tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\tan^2\dfrac{4\pi}{16}+\tan^2\dfrac{5\pi}{16}+\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}=$$ $$=\tan^2\dfrac{\pi}{16}+\cot^2\dfrac{\pi}{16}+\tan^2\dfrac{3\pi}{16}+\cot^2\dfrac{3\pi}{16}+\tan^2\dfrac{\pi}{8}+\cot^2\dfrac{\pi}{8}+1=$$ $$=\left(\tan\frac{\pi}{16}+\cot\frac{\pi}{16}\right)^2+\left(\tan\frac{3\pi}{16}+\cot\frac{3\pi}{16}\right)^2+\left(\tan\frac{\pi}{8}+\cot\frac{\pi}{8}\right)^2-5=$$ $$=\frac{1}{\sin^2\frac{\pi}{16}\cos^2\frac{\pi}{16}}+\frac{1}{\sin^2\frac{3\pi}{16}\cos^2\frac{3\pi}{16}}+\frac{1}{\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}}-5=$$ $$=\frac{4}{\sin^2\frac{\pi}{8}}+\frac{4}{\sin^2\frac{3\pi}{8}}+\frac{4}{\sin^2\frac{\pi}{4}}-5=$$

$$=\frac{4}{\sin^2\frac{\pi}{8}}+\frac{4}{\cos^2\frac{\pi}{8}}+3=\frac{4}{\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}}+3=\frac{16}{\sin^2\frac{\pi}{4}}+3=35$$

$\endgroup$
  • $\begingroup$ In your second line, the term that precedes $1$ should be $\cot^2\frac{\pi}{8}$. $\endgroup$ – N. F. Taussig Aug 6 '17 at 8:50
  • $\begingroup$ @N. F. Taussig It was typo. I fixed it. Thank you! $\endgroup$ – Michael Rozenberg Aug 6 '17 at 9:04
4
$\begingroup$

Use the following identity:

$$ \cot x - \tan x = 2 \cot 2x$$

Square this

$$ \cot^2 x + \tan^2 x = 2 + 4 \cot^2 2x$$

Use this again with your idea of grouping terms as $\tan^2 x + \cot^2x$, and you will see that all you need to know is the value of $$\cot \frac{4\pi}{16} = \cot \frac{\pi}{4}$$

$\endgroup$

Your Answer

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