1
$\begingroup$

It has been more than sixty years since I studied trigonometry, and I cannot come up with a fundamental proof that the product of cos(20), cos(40) and cos(80) is exactly 0.125. (Arguments in degrees, of course.) Can anyone help?

$\endgroup$
4
$\begingroup$

$$ \begin{split} \cos(20)\cos(40)\cos(80) &= \frac{\sin(20)\cos(20)\cos(40)\cos(80)}{\sin(20)} \\ &= \frac{0.5\sin(40)\cos(40)\cos(80)}{\sin(20)} \\ &= \frac{0.25\sin(80)\cos(80)}{\sin(20)} \\ &= \frac{0.125\sin(160)}{\sin(20)} \\ &= \frac{0.125\sin(20)}{\sin(20)} \\ &= 0.125 \\ \end{split} $$

$\endgroup$
  • $\begingroup$ Thanks so much. Perhaps I would have had more insight into this $\endgroup$ – user117232 Sep 24 '14 at 13:15
  • $\begingroup$ Thanks. Perhaps I would have had more insight into this one if the product had been shown as cos(A)cos(2A)cos(4A) $\endgroup$ – user117232 Sep 24 '14 at 13:19

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.