How to show $\tan \frac{\pi}{7} < \sin\frac{5\pi}{18}$ without a calculator?

Question

How can I show that $\tan \frac{\pi}{7} < \sin\frac{5\pi}{18}$? (without a calculator)

Attempts:

Since both sides are greater then zero, we can compare the squares of the functions.

Replaced $\tan \frac{\pi}{7}$ with $\cot (\frac{\pi}{2}-\frac{\pi}{7})=\cot\frac{5\pi}{14}$. Then I replaced $(\sin\frac{5\pi}{18})^2$ with $(\frac{1}{1+\cot \frac{5\pi}{18}})^2$. Then I tried to simplify the difference, but it didn't clarify anyhing. Could you please help me compare those functions?

Answer 1

Note that $\sin(\tfrac{5\pi}{18}) > \sin(\tfrac{\pi}{4}) = \tfrac{1}{\sqrt 2}$, and $\tan(\tfrac{\pi}{7})< \tan(\tfrac{\pi}{6})= \tfrac{1}{\sqrt 3}$.

Answer 2

• compare $\tan(\frac{\pi}7)$ to $\tan(\frac{\pi}6)$
• compare $\sin(\frac{5\pi}{18})$ to $\sin(\frac{\pi}4)$
• compare $\frac{\sqrt{2}}2$ to $\frac{\sqrt{3}}{3}$

You should be able to conclude.

Answer 3

Use that $$\sin\left(\frac{5\pi}{18}\right)-\tan\left(\frac{\pi}{7}\right)=-\sec\left(\frac{\pi}{7}\right)\left(\sin\left(\frac{\pi}{7}\right)-\cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{9}\right)\right)$$

Answer 4

It is not difficult since they are very distant from each other. The given inequality is equivalent to $$ \sin\tfrac{\pi}{7} < \sin\tfrac{5\pi}{18}\,\cos\tfrac{\pi}{7} $$ or to $$ 2\sin\tfrac{\pi}{7} < \sin\tfrac{53\pi}{126}+\sin\tfrac{17\pi}{126}$$ which can be proved through $$ 2\sin\tfrac{\pi}{7} < \tfrac{2\pi}{7} < \tfrac{10}{9}<\sin\tfrac{53\pi}{126}+\sin\tfrac{17\pi}{126}$$ since $18\pi<70$ and $\frac{2}{\pi}x<\sin x< x$ for $x\in\left(0,\frac{\pi}{2}\right)$.