Trigonometry inequation and induction

**Logic** · September 17th 2011, 04:43 AM

Hello,
I need to prove that |sinnx|<=n|sinx| using mathematical induction, but I get stuck as to how to use the inductive presumption.
Firstly, we need to check whether the given statement is true for n = 1.
We get |sinx|<=|sinx| which is, obviously, true.
Secondly, we presume that |sinnx|<=n|sinx| is true and we need to prove that it is also true for n+1, i.e. |sin(n+1)x|<=(n+1)|sinx|

The last inequation equals |sin(n+1)cosx + cos(n+1)sinx|<=(n+1)|sinx|

I am not certain how to proceed from here forth.
Do I need to use |x|-|y|<=|x+y|<=|x|+|y| - the triangle inequation to simplify somehow? And how does the presumption come into use?

**Prove It** · September 17th 2011, 05:06 AM

Originally Posted by Logic

Hello,
I need to prove that |sinnx|<=n|sinx| using mathematical induction, but I get stuck as to how to use the inductive presumption.
Firstly, we need to check whether the given statement is true for n = 1.
We get |sinx|<=|sinx| which is, obviously, true.
Secondly, we presume that |sinnx|<=n|sinx| is true and we need to prove that it is also true for n+1, i.e. |sin(n+1)x|<=(n+1)|sinx|

The last inequation equals |sin(n+1)cosx + cos(n+1)sinx|<=(n+1)|sinx|

I am not certain how to proceed from here forth.
Do I need to use |x|-|y|<=|x+y|<=|x|+|y| - the triangle inequation to simplify somehow? And how does the presumption come into use?

The triangle inequality is very useful, but you should first rewrite $\displaystyle \sin{[(n + 1)x]} = \sin{(nx + x)}$ . Then

$\displaystyle \begin{align*} |\sin{[(n+1)x]}| &= |\sin{(nx + x)}| \\ &= |\sin{(nx)}\cos{(x)} + \cos{(nx)}\sin{(x)}| \\ &\leq |\sin{(nx)}\cos{(x)}| + |\cos{(nx)}\sin{(x)}| \textrm{ by the triangle inequality} \\ &= |\sin{(nx)}||\cos{(x)}| + |\cos{(nx)}||\sin{(x)}| \\ &\leq n|\sin{(x)}||\cos{(x)}| + |\cos{(nx)}||\sin{(x)}| \textrm{ since }|\sin{(nx)}| \leq n|\sin{(x)}| \\ &= |\sin{(x)}|\left[n|\cos{(x)}| + |\cos{(nx)}|\right] \\ &\leq |\sin{(x)}|\left(n \cdot 1 + 1\right) \textrm{ since }|\cos{(x)}| \leq 1 \textrm{ and }|\cos{(nx)}| \leq 1 \\ &= (n + 1)|\sin{(x)}|\end{align*}$

Q.E.D.

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