Proving $\cos 36° > \tan 36° $ How do we  prove that  $\cos  36° > \tan 36° $ ? Please help . Thank you.

We all know that $\pi^2<10$. It follows that
$$\cos{\pi\over 5}>1-{1\over2}\left({\pi\over5}\right)^2>1-{10\over50}={4\over5}\ .$$
From this we conclude that
$$\cos^2{\pi\over5}>{32\over50}>{\pi\over5}>\sin{\pi\over5}\ ,$$
which is equivalent to the claim.
