MHB Solve Trigonometric Inequality 5x≤8sinx−sin2x≤6x

[lfdahl]

Show, that

$5x \le 8\sin x - \sin 2x \le 6x$ for $0 \le x \le \frac{\pi}{3}$.

 

[lfdahl]

Hint:

Spoiler

If $f(x) = 8\sin x - \sin 2x$, show that $5 \le f'(x) \le 6$ on $[0;\frac{\pi}{3}]$.

 

[lfdahl]

Suggested solution:

Spoiler

Let
\[f(x) = 8 \sin x - \sin 2x \\\\ f'(x) = 8 \cos x - 2 \cos 2x \\\\ f''(x) = -8 \sin x +4 \sin 2x = -8 \sin x(1- \cos x)\]

From these we see $f'(0) = 6, f'(\pi/3) = 5, f(0) = 0$ and $f''(x) \leq 0$ on $[0;\pi/3].$

Hence, $5\leq f'(x) \leq 6$ on $[0;\pi/3].$

Integrating from $0$ to $x$ gives

\[5x \leq f(x) \leq 6x.\]