Trigonometric inequality bounded by lines

[kalish1]

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]

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]

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]

We can show that this inequality is true on:

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

 

[CellCoree]

I would approach this problem by first understanding the context in which it arises - damped motion in spring-mass systems. This helps us to visualize the physical meaning behind the inequality and gives us a starting point for solving it.

Next, I would use my knowledge of trigonometric functions and their properties to simplify the equation. For example, we can use the identity $cos^2(x)+sin^2(x)=1$ to rewrite $16x\cos(8x)+4x\sin(8x)-2\sin(8x)$ as $16x\cos(8x)+2x\sin(8x)$. This simplification may help us to better understand the behavior of the inequality.

We can also use the fact that $|17x|=17x$ if $x\geq 0$ and $|17x|=-17x$ if $x<0$. This gives us two different cases to examine: when $x\geq 0$ and when $x<0$.

For the case of $x\geq 0$, we can use the properties of cosine and sine functions to show that $16x\cos(8x)+2x\sin(8x)$ is always less than or equal to $17x$. This can be seen by noting that the maximum value of $\cos(8x)$ is $1$ and the maximum value of $\sin(8x)$ is $1$ when $x=0$. Therefore, the left side of the inequality is always less than or equal to $17x$, which satisfies the inequality.

For the case of $x<0$, we can use similar reasoning to show that $16x\cos(8x)+2x\sin(8x)$ is always greater than or equal to $-17x$. This can be seen by noting that the minimum value of $\cos(8x)$ is $-1$ and the minimum value of $\sin(8x)$ is $-1$ when $x=0$. Therefore, the left side of the inequality is always greater than or equal to $-17x$, which also satisfies the inequality.

In conclusion, we have shown that for all values of $x$, the left side of the inequality is either less than or equal to $17x$ or greater than or equal to $-17x$, which satisfies the inequality. Therefore