prove by mathematical induction that for all positive integers n, sin(nπ+x)=(-1)^n
Base Case: When n = 1, we have sin(π + x) = -1, which is true since sin(π) = 0 and sin(x) = -1.
Inductive Step: Assume that sin(nπ + x) = (-1)^n is true for some positive integer n. We want to show that sin((n+1)π + x) = (-1)^(n+1).
Using the addition formula for sine, we have:
sin((n+1)π + x ) = sin(nπ + x + π)
= sin(nπ + x)cos(π) + cos(nπ + x)sin(π)
= sin(nπ + x)(-1) + cos(nπ + x)(0)
= -sin(nπ + x)
= -(-1)^n (by the inductive hypothesis)
= (-1)(-1)^n
= (-1)^(n+1)
Therefore, sin(nπ + x) = (-1)^n for all positive integers n by mathematical induction.