Show that $\ln(\sin(-20)+21)>3$ by hand Hi I hope this problem is new :
Show that :
$$\ln(\sin(-20)+21)>3$$
I have tried the power series of $\ln(x)$ and $\sin(x)$ without success because it's becomes hard by hand .
You can also find the inequality due to Michael Rozenberg wich states for $x\geq 1$ :
$$\ln(x)\leq (x-1)\left(\frac{2}{x^2+x}\right)^{\frac{1}{3}}$$
But unfortunately it's an upper bound not a lower bound so maybe we can play with it .
Edit :
We have :
$$k=\sin(-20)+21$$
Then it seems we have for $5\leq x\leq k$ :
$$\left(\frac{2x}{2+x}\left(1-b\right)+b\frac{\left(x-1\right)}{\sqrt{x}}\right)<\ln(x)$$
Where $b=0.484092$
I used the lower bound
Use Jensen's inequality to show $\frac{2x}{2+x} < \log(1+x) < \frac{2x+x^2}{2+2x}$ for $x>0$
And another well-know upper bound for the logarithm .
Question :
How to show it by hand (without a calculator) ?
 A: Let's approximate $\sin(-20)$ using a Taylor series centered near $-20$. Since $\pi \approx 3.14$, we have $-6.5\pi \approx -20.41$, so this is what we'll use as the center of the approximation. Note that $\sin(-6.5\pi) = -1$ and $\cos(-6.5\pi) = 0$. So we have
$$\sin(x) = \sum_{k=0}^{\infty} (-1)^k \frac{\sin(-6.5\pi)}{(2k)!} \cdot (x + 6.5\pi)^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(2k)!} (x + 6.5\pi)^{2k}.$$
We'll use the quadratic approximation:
$$\sin(x) = -1 + \frac{1}{2} \cdot (x + 6.5\pi)^2 + O((x + 6.5\pi)^4).$$
Since $3.14159 < \pi < 3.14160$, we have $0.42 < -20 + 6.5\pi < 0.421$. We compute
$$\sin(-6.5\pi + 0.42) \approx -1 + \frac{1}{2} \cdot 0.42^2 = -0.9118$$
and
$$\sin(-6.5\pi + 0.421) \approx -1 + \frac{1}{2} \cdot 0.421^2 = -0.9113795.$$
By Taylor's theorem, these approximations have error bounded above by $0.421^4/4! < 0.002$. So
$$-0.914 < \sin(-6.5\pi + 0.42) < \sin(-20) < \sin(-6.5\pi + 0.421) < -0.909.$$
Thus,
$$20.086 < \sin(-20) + 21 < 20.091.$$
At this point, we could try to approximate the natural logarithm, but its Taylor series converges rather slowly, so it's more efficient to bound $e^3$. We have $e < 2.7183$, so
$$e^3 < 2.7183^3 = 20.08593... < 20.086 < \sin(-20) + 21.$$
Thus $3 < \ln(\sin(-20) + 21)$.
A: Some thoughts:
Fact 1: $\sin 20 < \frac{21}{23}$.
Using Fact 1, it suffices to prove that
$\ln (21 - \frac{21}{23}) > 3$
or $\frac{462}{23} > \mathrm{e}^3$ which is true.


Proof of Fact 1:
The desired inequality is written as
$$\cos \frac{13\pi - 40}{2}  < \frac{21}{23}.$$
Using $\cos 2x = 1 - 2\sin^2 x$, it suffices to prove that
$$\sin \frac{13\pi - 40}{4} > \frac{1}{\sqrt{23}}.$$
Using $\sin y \ge y - \frac{1}{6}y^3$ for all $y \ge 0$, it suffices to prove that
$$\frac{13\pi - 40}{4} - \frac16 \left(\frac{13\pi - 40}{4}\right)^3
> \frac{1}{\sqrt{23}}$$
which is true.
