I think this goes in this section.
Let x be a real number. Prove that
abs(sin nx) <= n*abs(sinx) for all positive integers n.
I did not try to do the problem yet but you can use the fact that,Originally Posted by Nichelle14
$\displaystyle \sqrt{x^2}=|x|$
Thus, you need to prove that,
$\displaystyle \sqrt{\sin ^2 nx}\leq n\sqrt{\sin^2 x}$
If and only if,
$\displaystyle \sin^2 nx\leq n^2\sin ^2 x$
Mathematical induction is the tool here.
It is true for $\displaystyle n=1$ thus, there is a $\displaystyle k$ such as, $\displaystyle |\sin kx|\leq k|\sin x|$
Thus,
$\displaystyle 0\leq |\sin kx|\leq k|\sin x|$
But,
$\displaystyle 0\leq |\cos x|\leq 1$
Thus, (mutiply inequalities notice they are non-negative),
$\displaystyle |\sin kx||\cos x|\leq k|\sin x|$......(1)
Also,
$\displaystyle 0\leq |\sin x|\leq |\sin x|$
And,
$\displaystyle 0\leq |\cos kx|\leq 1$
Thus, (mutiply inequalites notice they are non-negative),
$\displaystyle |\cos kx||\sin x|\leq |\sin x|$......(2)
Using the property $\displaystyle |x||y|=|xy|$ on (1) and (2) we have,
$\displaystyle |\sin kx \cos x|\leq k|\sin x|$
$\displaystyle |\cos kx \sin x|\leq |\sin x|$
Now, add these inequalities,
$\displaystyle |\sin kx \cos x|+|\cos kx \sin x|\leq k|\sin x|+|\sin x|$
Now, by triangular inequality,
$\displaystyle |\sin kx \cos x|+|\cos kx \sin x|\geq |\sin kx \cos x+\cos kx \sin x|$
By transitivity ($\displaystyle a<b \mbox{ and }b<c \rightarrow a<c$),
$\displaystyle |\sin kx \cos x+\cos kx \sin x|\leq k|\sin x|+|\sin x|$
Thus, recognizing the sum for sine and simply the right side,
$\displaystyle |\sin (k+1) x|\leq (k+1)|\sin x|$
Proof is complete.
Of course.Originally Posted by Quick
$\displaystyle a\leq b$ is defined to be true whenever,
$\displaystyle a<b$ OR $\displaystyle a=b$.
Furthermore, in mathematics this expression is used a lot. There is one powerful theorem in set theory (Zorn's Lemma) which is based on the fact that $\displaystyle a\leq a $.