prove by mathematical induction that for all positive integers n, sin(nπ+x)=(-1)^n, using n=k and n=k+1 as proof

Answer 1

To prove that sin(nπ + x) = (-1)^n for all positive integers n, we will use mathematical induction.

Step 1: Base Case
For n = 1, we have sin(π + x) = (-1)^1, which simplifies to sin(π + x) = -1. This is true, as sin(π + x) = -sin(x) = -1.

Step 2: Inductive Hypothesis
Assume that for any positive integer k, sin(kπ + x) = (-1)^k.

Step 3: Inductive Step
We need to show that if the statement holds for k (i.e., sin(kπ + x) = (-1)^k), then it holds for k + 1 as well (i.e., sin((k + 1)π + x) = (-1)^(k + 1)).

Starting with the left-hand side:

sin((k + 1)π + x)
= sin(kπ + π + x) [using the property sin(a + b) = sin(a)cos(b) + cos(a)sin(b)]
= sin(kπ + x)cos(π) + cos(kπ + x)sin(π) [using the values of sin(π) = 0 and cos(π) = -1]
= sin(kπ + x)(-1) + cos(kπ + x)(0) [since sin(π) = 0 and cos(π) = -1]
= -sin(kπ + x)
= -(-1)^k [using the inductive hypothesis]
= (-1)(-1)^k
= (-1)^(k + 1).

Therefore, sin((k + 1)π + x) = (-1)^(k + 1), which proves the statement for n = k + 1.

By the principle of mathematical induction, the statement is true for all positive integers n.