Trigonometry | PRMO-2018 | Problem No-14

Pre College Mathematics

Prmo-2018, Problem-14

\(19\)

Given that $x=\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \cdots \cos 89^{\circ}$

\(\Rightarrow x=\cos 1^{\circ}\cos 89^{\circ}\cos 2^{\circ}\cos 88^{\circ}........\cos 45^{\circ}\)

\(\Rightarrow x=\cos 1^{\circ}\sin 1^{\circ}\cos 2^{\circ}\cos 2^{\circ}........\cos 45^{\circ}\) ( as cos(90-x)=sin x)

\(\Rightarrow x=\frac{1}{2^{44}} (2 \cos 1^{\circ} \sin 1^{\circ}) (2\cos 2^{\circ}\sin 2^{\circ})........\cos 45^{\circ}\)

$\Rightarrow x =\frac{ \cos 45^{\circ} \cdot \sin 2^{\circ} \cdot \sin 4^{\circ} \cdots \sin 88}{ 2^{44}}$ (as sin 2x= 2 sin x cos x)

Now can you finish the problem?

Therefore we can say that
$x=\frac{1}{2^{1 / 2}} \times \frac{\sin 4^{\circ} \cdot \sin 8^{\circ} \cdots \sin 88^{\circ}}{2^{44} \times 2^{22}}$
$=\frac{\sin \left(90-86^{\circ}\right) \cdot \sin \left(90-84^{\circ}\right) \cdots \sin (90-2)}{2^{66} \cdot 2^{1 / 2}}$
$=\frac{\cos 2^{\circ} \cdot \cos 6^{\circ} \cdot \cos 86^{\circ}}{2^{133 / 2}}$
$x=\frac{y}{2^{133 / 2}}$

Can you finish the problem...?

$\frac{y}{x}=2^{133 / 2}$
$ \frac{2}{7} \log \left(\frac{y}{x}\right)=\frac{2}{7} \times \log _{2}(2)^{133 / 2}$
$=\frac{2}{7} \times \frac{133}{2}$
$=19$