Is Sin 1 Less Than Log Base 3 of Root 7?

[anemone]

Here is this week's POTW:

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Prove $\sin 1 < \log_3 \sqrt{7}$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to kaliprasad for his correct solution, which you can find below::)

Spoiler

We have $\sin \,1 < \sin \frac{\pi}{3} $ as $ 1 < \frac{\pi}{3}$
or $\sin \,1 < \frac{\sqrt{3}}{2}\cdots (1)$

Note that $(4\sqrt{3})^2 = 48 < 49 = 7^2$, hence $\sqrt{3} < \frac{7}{4}$

Combining the results we get
$\sin \,1 < \frac{7}{8}\cdots (2)$

Now observe that $3^7 = 2187$ and $7^4 = 2401$ hence

$3^7 < 7^4$

$3^\frac{7}{8} < \sqrt{7}$

$\frac{7}{8} < \log_3\sqrt{7}$

from (2) and above we have proved that $\sin\,1 < log_3 \sqrt{7}$.