# Trig

• Nov 30th 2010, 01:43 PM
chiph588@
Trig
Show $\displaystyle \tan \frac\pi9 + 4\sin \frac\pi9 = \sqrt3$.
• Dec 1st 2010, 12:33 AM
simplependulum
Quote:

Originally Posted by chiph588@
Show $\displaystyle \tan \frac\pi9 + 4\sin \frac\pi9 = \sqrt3$.

Consider $\frac{\pi}{3} - \frac{\pi}{9} = \frac{2\pi}{9}$

$\sin(\frac{\pi}{3} - \frac{\pi}{9}) = \sin(\frac{2\pi}{9})$

$\frac{\sqrt{3}}{{2}} \cos(\frac{\pi}{9}) - \frac{1}{2} \sin(\frac{\pi}{9}) = 2\sin(\frac{\pi}{9}) \cos(\frac{\pi}{9})$

$\sqrt{3} \cos(\frac{\pi}{9}) - \sin(\frac{\pi}{9}) = 4 \sin(\frac{\pi}{9}) \cos(\frac{\pi}{9})$

$\sqrt{3} = \tan(\frac{\pi}{9}) + 4 \sin(\frac{\pi}{9})$
• Dec 1st 2010, 10:08 AM
chiph588@
Quote:

Originally Posted by simplependulum
Consider $\frac{\pi}{3} - \frac{\pi}{9} = \frac{2\pi}{9}$

$\sin(\frac{\pi}{3} - \frac{\pi}{9}) = \sin(\frac{2\pi}{9})$

$\frac{\sqrt{3}}{{2}} \cos(\frac{\pi}{9}) - \frac{1}{2} \sin(\frac{\pi}{9}) = 2\sin(\frac{\pi}{9}) \cos(\frac{\pi}{9})$

$\sqrt{3} \cos(\frac{\pi}{9}) - \sin(\frac{\pi}{9}) = 4 \sin(\frac{\pi}{9}) \cos(\frac{\pi}{9})$

$\sqrt{3} = \tan(\frac{\pi}{9}) + 4 \sin(\frac{\pi}{9})$

Much more elegant than my solution!
• Dec 1st 2010, 11:56 AM
TheCoffeeMachine
Quote:

Originally Posted by chiph588@
Much more elegant than my solution!

What was your solution? De Moivre/binomial/Vieta?
• Dec 4th 2010, 08:16 PM
chiph588@
Quote:

Originally Posted by TheCoffeeMachine
What was your solution? De Moivre/binomial/Vieta?

Let $u = \cos\left(\frac\pi9\right)$. From the third angle formula, we obtain

$8u^3-6u-1 = 0$

$-(2u+1)(8u^3-6u-1) = 0$

$-16u^4-8u^3+12u^2+8u+1 = 0$

$(1-u^2)(1+4u)^2 = 3u^2$

$\sqrt{1-u^2}(1+4u) = \sqrt3u$

$\sqrt{1-u^2}+4u\sqrt{1-u^2} = \sqrt3u$

$\sin\left(\frac\pi9\right)+4\cos\left(\frac\pi9\ri ght)\sin\left(\frac\pi9\right) = \sqrt3\cos\left(\frac\pi9\right)$

$\tan\left(\frac\pi9\right)+4\sin\left(\frac\pi9\ri ght) = \sqrt3$
• Dec 5th 2010, 02:57 PM
chiph588@
Quote:

Originally Posted by chiph588@
Let $u = \cos\left(\frac\pi9\right)$. From the third angle formula, we obtain

$8u^3-6u-1 = 0$

$-(2u+1)(8u^3-6u-1) = 0$

$-16u^4-8u^3+12u^2+8u+1 = 0$

$(1-u^2)(1+4u)^2 = 3u^2$

$\sqrt{1-u^2}(1+4u) = \sqrt3u$

$\sqrt{1-u^2}+4u\sqrt{1-u^2} = \sqrt3u$

$\sin\left(\frac\pi9\right)+4\cos\left(\frac\pi9\ri ght)\sin\left(\frac\pi9\right) = \sqrt3\cos\left(\frac\pi9\right)$

$\tan\left(\frac\pi9\right)+4\sin\left(\frac\pi9\ri ght) = \sqrt3$

If you're wondering how I came up with this, it's called working backwards (Tongueout).