One problem to solve

[Polyche]

Let Sn= sinx x sin2x x sin3x x...x sin nx=1
S100=?

 

[lev888]

Let Sn= sinx x sin2x x sin3x x...x sin nx=1
S100=?
View attachment 29148

In the image I don't see that Sn = 1, as you wrote. Please clarify.

 

[lex]

[imath]S_1[/imath] is the solution set of the equation [imath]\sin{x}=1[/imath]
This equation can easily be solved.

[imath]S_{100}[/imath] is the solution set of the equation [imath]\prod \limits_{k=1}^{100} \sin{kx} = 1[/imath]
If you consider when this product would be 1, you will see that there are no (real) solutions.

 

[Steven G]

If I answer based on what you wrote, then the answer is clearly 1.

Please do not use x to denote multiplication and the variable x.

You wrote Sn= sinx*sin2x*sin3x*...* sin nx=1. This means for all n, the left hand side equals 1. So S100 = 1

 

[Dr.Peterson]

Let Sn= sinx x sin2x x sin3x x...x sin nx=1
S100=?
View attachment 29148

You have caused confusion by not translating the entire problem into English. The answer by lex clarifies the problem; S_n is not the expression you say it is equal to, but the solution set of the equation that follows. You should have written

Let S_n be the solution set of the equation sin x sin 2x sin 3x ... sin nx = 1.​

If I were you, I would have found S₁, then tried S₂, and then realized what the key idea is for any larger n: What is the greatest possible value of each of the factors? What is required to make the product equal to 1?