How do we prove that $\cos 36° > \tan 36° $ ? Please help . Thank you.
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$\begingroup$Since $\cos x$ and $\tan x$ are monotonic in the $[0,\pi/2]$, they cross at a single point, namely when: $$\cos^2x=\sin x\ \rightarrow\ 1 -\sin^2x-\sin x =0 $$ Or when: $$\sin x=\varphi-1\ \rightarrow \ x\approx38.2^\circ$$ Since $36^\circ<38.2^\circ$, we also have $\cos 36^\circ >\tan 36^\circ$. $_\blacksquare$
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$$\cos36 > \tan36 \iff \cos36 > \dfrac{\sin36}{\cos36} \iff \cos^2{36} > \sin36 \iff 1 - \sin^2{36} > \sin36 \iff \sin^2{36} + \sin36 - 1 < 0 \iff \sin36 < \dfrac{\sqrt{5} - 1}{2} (*)$. We prove $(*)$ true. let $x = \sin18$, then using $\cos36 = \sin54$ gives: $\cos(2\cdot 18) = \sin(3\cdot 18) \iff 1 - 2x^2 = 3x - 4x^3 \iff (x - 1)(4x^2 + 2x -1) = 0$. Since $0 < x < 1$, this equation gives only one solution: $x = \dfrac{\sqrt{5} - 1}{4}$. So $\cos36 = 1- 2x^2 = \dfrac{\sqrt{5} + 1}{4}$. So $\sin36 = \sqrt{1 - \cos^2{36}} = \dfrac{\sqrt{10 - 2\sqrt{5}}}{4} < \dfrac{\sqrt{5} - 1}{2} \iff \dfrac{10 - 2\sqrt{5}}{16} < \dfrac{6 - 2\sqrt{5}}{4} \iff 3\sqrt{5} < 7 \iff 45 < 49$. The last inequality is obviously true. This implies $(*)$ is true, and we're done.
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