Thread: Cos and Tan

1. Cos and Tan

Prove : $\cos 36^\circ > \tan 36^\circ$ $
$

2. $\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$

$\sin 36^{\circ}=\frac{(\sqrt{5}-1)\sqrt{1+2\sqrt{5}}}{8}$

Then $\tan 36^{\circ}=\frac{(3-\sqrt{5})\sqrt{10+2\sqrt{5}}}{4}$

We have to prove $\sqrt{5}+1>(3-\sqrt{5})\sqrt{10+2\sqrt{5}}$

Square both members:

$6+2\sqrt{5}>(14-6\sqrt{5})(10+2\sqrt{5})$

$17\sqrt{5}>37$

Square again both members: $1440>1369$ which is true.

3. cos A = tan A

cos A - sin A / cos A = 0

cos^2 A - sin A = 0

(1 - sin^2 A) - sin A = 0

sin^2 A + sin A - 1 = 0

sin A = (-1 +/- (1 - 4(1)(11))^(1/2)/[2(1)]

sin A = (-1 +/- sqrt 5)/2

A = 45 degress = pi/4 rad

that means if A = 0 to A<pi/4, cos A > tan A