# grrr, this assn is giving me a lot of troubble 2 more ?, and then I'll be done

• December 17th 2007, 04:12 AM
Luckegrl_16
grrr, this assn is giving me a lot of troubble 2 more ?, and then I'll be done
1) If tan (theta) = 1/2, the value of tan (4pi/3 - (theta)) = a (sq.root of b) - c. Find a, b, and c.

THANKS... again, for all the help :D
• December 17th 2007, 04:37 AM
TKHunny
0) No one wants to do your homework for you. If you show NO WORK AT ALL, this is what it seems we will be doing if we solve problems for you.

1) You should have a formula in your book for the tangent of a sum of angles. Also, you simply should KNOW tan(4pi/3)

2) It may be time for you to start plotting some points - unless you have a sufficiently capable calculator.

• December 17th 2007, 05:04 AM
Soroban
Hello, Luckegrl_16!

Quote:

$\text{1) If }\tan\theta = \frac{1}{2}\text{, the value of: }\:\tan\left(\frac{4\pi}{3} - \theta\right) \;=\;a\sqrt{b} \;-\;c.\qquad\text{Find }a,\,b\text{, and }c.$
We're expected to know: . $\tan(A - B) \:=\:\frac{\tan A - \tan B}{1 + \tan A\tan B}$

Then: . $\tan\left(\frac{4\pi}{3} - \theta\right) \;=\;\frac{\tan\frac{4\pi}{3} - \tan\theta}{1 + \tan\frac{4\pi}{3}\tan\theta} \;=\;\frac{\sqrt{3} - \frac{1}{2}}{1 + \sqrt{3}\!\cdot\!\frac{1}{2}} \;=\;\frac{2\sqrt{3}-1}{2 + \sqrt{3}}$

Rationalize: . $\frac{2\sqrt{3}-2}{2 + \sqrt{3}}\cdot{\color{blue}\frac{2-\sqrt{3}}{2-\sqrt{3}}} \;=\;\frac{4\sqrt{3} - 6 - 2 + \sqrt{3}}{4-3} \;=\;\frac{-8+5\sqrt{3}}{1} \;=\;5\sqrt{3} - 8$

$\text{Therefore: }\:\boxed{\;a = 5,\;\;b = 3,\;\;c = 8\;}$

• December 17th 2007, 05:11 AM
Luckegrl_16
well, I knew the formula, and tan 4pi/3, and other little details, but i really was confused on how to put it all together... THANKS Soroban!!
• December 17th 2007, 07:23 AM
TKHunny
That really hurts.

In other words, you still haven't shown any work. Please step up your game or you will not be learning much.