Let S = sin(π+x)+ sin(2π+x)+sin(3π+x)+...+sin(nπ+x) for 0<x<π/2 and all positive integers n. Show that -1<S<=0
We can rewrite S as sin(π+x)+sin(2π+x)+sin(3π+x)+...+sin(nπ+x) for 0<x<π/2 and all positive integers n.
Now, let's consider the maximum value of S. Since sin(x) has a maximum value of 1, the sum of sin(π+x)+sin(2π+x)+sin(3π+x)+...+sin(nπ+x) cannot exceed n.
Therefore, the maximum value of S is n. Since n is a positive integer, the maximum value of S is positive.
Now, let's consider the minimum value of S. Since sin(x) has a minimum value of -1, the sum of sin(π+x)+sin(2π+x)+sin(3π+x)+...+sin(nπ+x) cannot be less than - n.
Therefore, the minimum value of S is - n. Since n is a positive integer, the minimum value of S is negative.
Therefore, - n < S ≤ n for all positive integers n.
Since n is positive, we have -1 ≤ S ≤ 0.
Therefore, we have shown that -1<S≤0.