# Trigonometric inequality $\sin(\pi x)>\cos(\pi \sqrt x)$

Solve the inequality

$\sin(\pi x)>\cos(\pi \sqrt x)$

I don't know where to begin. Hints?

• The inequality holds in a union of infinite disjoint intervals which is quite hard to describe. Where does this inequality come from? – Robert Z Oct 25 '16 at 6:29
• Didn't you forget another square root ? – Yves Daoust Oct 25 '16 at 6:32
• The sign of $\sin(\pi x)>\cos(\pi \sqrt x)$ changes infinitely many times so the inequality is false in general. – Nilotpal Sinha Oct 25 '16 at 6:39
• It's from my tutor, oh then it must be a typo or sth. – lmc Oct 25 '16 at 6:40

Thought:

Write $\sin (\pi x ) - \cos( \pi \sqrt{x} ) > 0$. Note $\sin ( \pi x) = \cos ( \pi/2 - \pi x) = \cos \left( \frac{ \pi - 2 \pi x }{2} \right)$. Thus,

$$\cos \left( \frac{ \pi - 2 \pi x }{2} \right) - \cos( \pi \sqrt{x} ) = - 2 \sin \left( \frac{ \frac{\pi - 2 \pi x}{2} + \pi \sqrt{x} }{2} \right)\sin \left( \frac{ \frac{\pi - 2 \pi x}{2} - \pi \sqrt{x} }{2} \right) =$$

$$= - 2 \sin \left( \frac{ \pi ( 1 - 2x + \sqrt{x} ) }{4} \right) \sin \left( \frac{ \pi ( 1 - 2x - \sqrt{x} ) }{4} \right)$$

And this is at least $0$ iff

$$\sin \left( \frac{ \pi ( 1 - 2x + \sqrt{x} ) }{4} \right) \sin \left( \frac{ \pi ( 1 - 2x - \sqrt{x} ) }{4} \right) < 0$$

• @Now_now_Draco_play_nicely, Could please the rest in case you've completed it – lab bhattacharjee Oct 25 '16 at 17:54

Hint:

The difference between the two members is a doubly oscillating function that stays in range $[-2,2]$. The oscillations due to the RHS get slower and slower.

There is no $x$ for which the two members both equal $1$ or $-1$ (by the transcendence of $\pi$), so that all the roots are simple.

Hence, the inequality is true for all $x$ between every other pair of roots, starting from the first pair (for $x>0$).

From the work of @ILoveMath, you can see that by means of the sum-to-product formula, you can compute those roots by solving quadratic equations in $\sqrt x$.