Trigonometric inequality bounded by lines

Question

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:

Thanks!

Answer 1

Note: Your result is not true!

If $x=-0.35$ then left hand side is $\approx 6.415$ and right hand side is $5.95$.

However, the following shows that you are almost right. Also, if you use $21|x|$ then your bound is valid. $17$ is not big enough.

For the rest of the analysis:

You know that $$ |A \cos(\theta) + B\sin(\theta)| \le \sqrt{A^2+B^2} $$ Working with squares $$ \text{LHS}^2 \le 256 x^2 + (4x-2)^2 = 272 x^2 -16 x + 4 \\ \text{RHS}^2 = 289 x^2 $$ This shows that for $|x|$ sufficiently large the inequality holds. Solving LHS=RHS we get $$ 272 x^2 -16 x + 4 = 289 x^2 \Rightarrow x=-{{2\,\sqrt{33}+8}\over{17}} , x={{2\,\sqrt{33}-8}\over{17 }} $$ Inside that region, look at the derivative of the left hand side and do it separately for $x>0$ and $x<0$.

P.S. I don't like my answer. I wish I had a more elegant solution. At least it shows that outside the interval $\approx [-1.15, 0.21]$, the result holds.

Answer 2

Since $\sin(x)$ and $\cos(x)$ are bounded from above by their coefficients, I would actually suppose that $$16x\cos(8x)+4x\sin(8x)-2\sin(8x)<|20x-2|$$

If you look closely, $|17x|$ actually seems to intersect your sinusoidal graph around $x=.5$.

You may be able to show that your function is bounded by $|17x|$ for sufficiently large $x$, but not for all.

Answer 3

The points of interest are the extrema, so take the derivative and set it equal to zero. Check at each extremum that your condition is met.