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
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
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.
Please give some indication of your level of understanding.
Hello, Luckegrl_16!
We're expected to know: .$\displaystyle \tan(A - B) \:=\:\frac{\tan A - \tan B}{1 + \tan A\tan B}$$\displaystyle \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.$
Then: .$\displaystyle \tan\left(\frac{4\pi}{3} - \theta\right) \;=\;\frac{\tan\frac{4\pi}{3} - \tan\theta}{1 + \tan\frac{4\pi}{3}\tan\theta} \;=\;\frac{\sqrt{3} - \frac{1}{2}}{1 + \sqrt{3}\!\cdot\!\frac{1}{2}} \;=\;\frac{2\sqrt{3}-1}{2 + \sqrt{3}}$
Rationalize: .$\displaystyle \frac{2\sqrt{3}-2}{2 + \sqrt{3}}\cdot{\color{blue}\frac{2-\sqrt{3}}{2-\sqrt{3}}} \;=\;\frac{4\sqrt{3} - 6 - 2 + \sqrt{3}}{4-3} \;=\;\frac{-8+5\sqrt{3}}{1} \;=\;5\sqrt{3} - 8$
$\displaystyle \text{Therefore: }\:\boxed{\;a = 5,\;\;b = 3,\;\;c = 8\;}$