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Trigonometry Trigonometric inequality bounded by lines

kalish

Member
Oct 7, 2013
99
How can it be shown that $$16x\cos(8x)+4x\sin(8x)-2\sin(8x)<|17x|?$$

This problem arises from work with damped motion in spring-mass systems in Differential Equations. I have gotten to this inequality after some algebraic manipulation, but am completely stuck here.

Here is the illustrative graph provided by Wolfram Alpha:

[1]: http://i.stack.imgur.com/oWf9E.png

Thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would use a linear-combination identity to obtain the amplitude $A$ of the trigonometric expression:

\(\displaystyle A=\sqrt{(16x)^2+(4x-2)^2}<\sqrt{(17x)^2}\)

What do you find?
 

lfdahl

Well-known member
Nov 26, 2013
722
The LHS can be expressed as a vector dot product:

\[16xcos(8x)+4xsin(8x)-2sin(8x)=16xcos(8x)+(4x-2)sin(8x)=\\\\ \binom{16x}{4x-2}\cdot \binom{cos(8x)}{sin(8x)}=\vec{a}\cdot \vec{e}\\\\ Applying\; the \; Cauchy-Schwarz \; inequality:\\\\ \left |\binom{16x}{4x-2}\cdot \binom{cos(8x)}{sin(8x)} \right |\leq \left \| \binom{16x}{4x-2} \right \|=\sqrt{(16x)^{2}+(4x-2)^2}\leq \sqrt{(17x)^{2}}\]

My question: How do you show the validity of the last inequality, if we don´t know the range of x:

\[\sqrt{(16x)^{2}+(4x-2)^2}\leq \sqrt{(17x)^{2}}\;?\]
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We can show that this inequality is true on:

\(\displaystyle \left(-\infty,\frac{2\left(-4-\sqrt{33} \right)}{17} \right)\,\cup\,\left(\frac{2\left(-4+\sqrt{33} \right)}{17},\infty \right)\)