Trigonometric inequality bounded by lines 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!
 A: 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.
A: 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.
A: 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.
