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.

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- Jul 5th 2006, 04:02 PMNichelle14Proof
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. - Jul 5th 2006, 04:41 PMThePerfectHackerQuote:

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,

$\displaystyle \sin^2 nx\leq n^2\sin ^2 x$ - Jul 8th 2006, 05:25 PMNichelle14
I still don't understand what to do.

- Jul 8th 2006, 07:42 PMThePerfectHacker
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. - Jul 9th 2006, 03:58 AMQuick
Is it possible to say $\displaystyle 5\leq5$

(because it seems somewhat ridiculous) - Jul 9th 2006, 05:47 AMThePerfectHackerQuote:

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 $.