Trigonometry! #144

We are given that $\cos(\theta) = \tan(\theta) \Longrightarrow \cos^{2}(\theta) = \sin(\theta)$

$\Longrightarrow 1 - \sin^{2}(\theta) = \sin(\theta) \Longrightarrow \sin^{2}(\theta) + \sin(\theta) - 1 = 0$

$\sin(\theta) = \dfrac{-1 \pm \sqrt{5}}{2}$ ,

but as $|\sin(\theta)| \le 1$ we have the unique solution $\sin(\theta) = \dfrac{\sqrt{5} - 1}{2}.$

Now $\sin(18^{\circ}) = \dfrac{\sqrt{5} - 1}{4}$ , so $n = \boxed{2}.$

Comments: To calculate $\sin(18^{\circ})$ , note that, with $t = \sin(18^{\circ})$ , we have that

$\sin(54^{\circ}) = \sin(3*18^{\circ}) = 3\sin(18^{\circ}) - 4\sin^{3}(18^{\circ}) = 3t - 4t^{3}$ and

$\cos(36^{\circ}) = \cos(2*18^{\circ}) = 1 - 2\sin^{2}(18^{\circ}) = 1 - 2t^{2}.$

But $\sin(54^{\circ}) = \cos(36^{\circ})$ , so $3t - 4t^{3} = 1 - 2t^{2}$

$\Longrightarrow 4t^{3} - 2t^{2} - 3t + 1 = 0 \Longrightarrow (t - 1)(4t^{2} + 2t - 1) = 0.$

Now clearly $t \ne 1$ , so we must have $t = \dfrac{-2 \pm \sqrt{4 + 16}}{8} = \dfrac{-1 \pm \sqrt{5}}{4}.$

Clearly $t = \sin(18^{\circ}) \gt 0$ , so we end up with $t = \sin(18^{\circ}) = \dfrac{\sqrt{5} - 1}{4}.$

$18^\circ \times 5=90^\circ$ then if $\alpha=18^\circ$ , $sin 2\alpha=cos 3\alpha$ $\leftrightarrow 2sin \alpha.cos \alpha=4cos^{3}\alpha-3cos\alpha$ $\leftrightarrow 2sin \alpha=4cos^{2}\alpha-3$ $\leftrightarrow 2sin \alpha=1-4sin^{2}\alpha$

Put $w=2sin \alpha$ , then $w=1-w^2$

Back to the origin:

$cos \theta=tan \theta \leftrightarrow cos^{2}\theta=sin \theta$

$\leftrightarrow 1-sin^{2}\theta=sin\theta$

which means $w=sin \theta$ . Therefore $sin \theta = 2sin \alpha$