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

Answer 1

First, let's establish the base case when n = 1.

When n = 1, we have sin(1π + x) = sin(π + x). By sine addition formula, we can rewrite this as:

sin(π)cos(x) + cos(π)sin(x) = 0⋅cos(x) + (-1)⋅sin(x) = -sin(x)

Therefore, the statement holds for the base case when n = 1.

Now let's assume the statement holds for some positive integer k, i.e., sin(kπ + x) = (-1)^k sin(x). This is our induction hypothesis.

Now, we need to prove that the statement also holds for k + 1. In other words, we want to prove that sin((k + 1)π + x) = (-1)^(k + 1) sin(x).

Starting with the left-hand side:

sin((k + 1)π + x)

Using the sine addition formula, we can rewrite this as:

sin(kπ + π + x)

Now, we can apply our induction hypothesis:

sin(kπ + π + x) = (-1)^k sin(π + x)

Using the sine addition formula again, we have:

(-1)^k sin(π)cos(x) + cos(π)sin(x)

Since sin(π) = 0 and cos(π) = -1, this simplifies to:

0⋅cos(x) + (-1)⋅sin(x) = -sin(x)

Which is equal to:

(-1)^(k + 1) sin(x) by the definition of the exponentiation with negative base.

Therefore, we have shown that if the statement is true for k, then it is also true for k + 1.

By the principle of mathematical induction, the statement "sin(nπ + x) = (-1)^n sin(x)" is proven for all positive integers n.