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

1. ## 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

2. 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.

3. Hello, Luckegrl_16!

$\text{1) If }\tan\theta = \frac{1}{2}\text{, the value of: }\:\tan\left(\frac{4\pi}{3} - \theta\right) \;=\;a\sqrt{b} \;-\;c.\qquad\text{Find }a,\,b\text{, and }c.$
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: . $\frac{2\sqrt{3}-2}{2 + \sqrt{3}}\cdot{\color{blue}\frac{2-\sqrt{3}}{2-\sqrt{3}}} \;=\;\frac{4\sqrt{3} - 6 - 2 + \sqrt{3}}{4-3} \;=\;\frac{-8+5\sqrt{3}}{1} \;=\;5\sqrt{3} - 8$

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

4. 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!!

5. 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.