prove by mathematical induction that for all positive integers n, sin(nπ+x)=(-1)^n sin x
First, we will prove the base case where n = 1:
sin(1π + x) = (-1)^1 sin x
sin(π + x) = -sin x
Since sin(π) = 0, we have -sin x = -sin x.
Now, assume that the statement is true for some positive integer k:
sin(kπ + x) = (-1)^k sin x
We will prove that the statement is also true for k + 1:
sin((k + 1)π + x) = (-1)^(k + 1) sin x
Using the sum of angles formula for sine, we can write:
sin((k + 1)π + x) = sin(kπ + x + π)
= sin(kπ + x)cos(π) + cos(kπ + x)sin(π)
= sin(kπ + x)(-1) + cos(kπ + x)(0) [since cos(π) = -1 and sin(π) = 0]
= -sin(kπ + x)
= -(-1)^k sin x [using the induction hypothesis]
= (-1)^1 * (-1)^k sin x
= (-1)^(k + 1) sin x
Therefore, the statement is proven for k + 1. By mathematical induction, the statement is true for all positive integers n.