Answer to Question #264609 in Real Analysis for Nikhil Singh

Question

Use the principle of mathematical induction to show that

| sin nx| ≤ n| sin x|

for all n∈ N and for all x ∈ R

Answer 1

Using the principle of mathematical induction let us show that

$| \sin nx| ≤ n| \sin x|$

for all $n∈ \N$ and for all $x ∈ \R.$

If $n=1$ then the inequality $| \sin x| ≤ | \sin x|$ is true for all $x ∈ \R.$

Suppose for $n=k$ that the inequality $| \sin kx| ≤ k| \sin x|$ is true for all $x ∈ \R.$

Let us prove the inequality for $n=k+1:$

$|\sin((k+1)x)|=|\sin(kx+x)|=|\sin(kx)\cos x+\cos(kx)\sin x|$

$\le |\sin(kx)|\cdot|\cos x|+|\cos(kx)|\cdot|\sin x| \le |\sin(kx)|\cdot1+1\cdot|\sin x|$

$\le k|\sin x|+|\sin x|=(k+1)|\sin x|.$

We conclude that according to the principle of mathematical induction the inequality $| \sin nx| ≤ n| \sin x|$ is true for all $n∈ \N$ and for all $x ∈ \R.$