For Trigo lovers-1

Geometry Level 4

$\large \sin( x) \sin( 2x) \sin( 3x ) \sin( 4x)=\frac{3}{4}$

Find number of real solutions of the above equation in the interval $(0,2015\pi)$ .

If number of solutions are infinite enter 12345 as your answer.

This problem belongs to the set: Trigo Plus! .

The answer is 0.

1 solution

Ravi Dwivedi
Jul 13, 2015

$\begin{aligned} & \sin x \sin 2x \sin 3x \sin 4x\\ =& \frac{1}{4}(\cos 3x - \cos 5x)(\cos x - \cos 5x)\\ =&\frac{1}{4}(\cos^{2} 5x - \cos 3x \cos 5x -\cos 5x \cos x + \cos x \cos 3x)\\ =&\frac{1}{8}(2 \cos^{2} 5x - \cos 2x + \cos 8x - \cos 4x + \cos 6x \\ < &\frac{1}{8}(2+1+1+1) = \frac{6}{8}= \frac{3}{4} \end{aligned}$

Hence the equation has no real solutions

Moderator note:

The maximum of the product is about $0.37$ .

How much more can we improve on $\frac{3}{4}$ ? Can we do $\frac{1}{2}$ ?

Hint: Use complex numbers to understand the product of sines, which allow us to easily arrive at another formula.

It's a beautiful question but the answer can be deduced with no effort. There is either going to be $0$ or an infinite amount of solutions since it's periodic. The way the question is posed it is obviously $0$ . Maybe you should change the question so it includes a range of values $x$ can be.

Isaac Buckley - 5 years, 11 months ago

Thanks for pointing out I have edited

Ravi Dwivedi - 5 years, 11 months ago

I'm pretty sure the 4th line is wrong.

Joe Mansley - 1 year, 7 months ago

For Trigo lovers-1

This problem belongs to the set: Trigo Plus! .

The answer is 0.

1 solution

Moderator note:

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