Please help. WILL MEDAL! Trigonometry @TheSmartOne @myininaya @nincompoop @jigglypuff314 @johnweldon1993 @jim_thompson5910 @mathstudent55 @zepdrix @Zarkon @calculusxy @iPwnBunnies

Question

okdutchman7 · Answer

$$3\cot \frac{ \theta }{ 2 } -\sqrt{3} = 0$$

okdutchman7 · Answer

$$3\cot \frac{ \theta }{ 2 } = \sqrt{3}$$

mathstudent55 · Answer

Divide both sides by 3.

okdutchman7 · Answer

Do I just divide the argument of the cotangent function or the whole thing?

okdutchman7 · Answer

$$\cot \frac{ \theta }{ 2 } = \frac{ \sqrt{3} }{ 3 }$$

mathstudent55 · Answer

Divide both sides just by 3.
This does not involve the argument of the cotangent.
For example:

$\dfrac{4 \tan \frac{\theta}{2}}{4} = \tan \frac{\theta}{2}$

okdutchman7 · Answer

Is that ^ right then?

mathstudent55 · Answer

$3\cot \frac{ \theta }{ 2 } -\sqrt{3} = 0$

$3\cot \frac{ \theta }{ 2 } = \sqrt{3} $

$\cot \frac{ \theta }{ 2 } = \dfrac{\sqrt{3}}{3} $

This is correct.
The division by 3 is not in the argument. It is only in the coefficient.

okdutchman7 · Answer

What do I do next?

mathstudent55 · Answer

The cotangent is the reciprocal of the tangent.

$\tan \frac{ \theta }{ 2 } = \dfrac{3}{\sqrt{3}} = \sqrt 3 $

okdutchman7 · Answer

tan is opposite/agacent

mathstudent55 · Answer

Now we solve for $\dfrac{\theta}{2} $.

For what value of an angle is the tangent equal to $\sqrt 3$ ?

okdutchman7 · Answer

The angle is 60

okdutchman7 · Answer

are there others?

mathstudent55 · Answer

|dw:1463707657811:dw|

mathstudent55 · Answer

Are you solving in an interval, or you need all values?

okdutchman7 · Answer

I need all values between 0 and 360.

mathstudent55 · Answer

|dw:1463707849929:dw|

mathstudent55 · Answer

The tangent is positive in quadrants I and III.

okdutchman7 · Answer

so 240

okdutchman7 · Answer

Is the $$\tan \frac{ \theta }{ 2 }$$ theta over 2 significant?

mathstudent55 · Answer

We solved so far for $\dfrac{\theta}{2} $, and we get that the tangent is $\sqrt 3 $ at 
$60^\circ + 180^\circ k$

We now have

$\dfrac{\theta}{2} = 60^\circ + 180^\circ k$

Now we solve for $\theta$ by multiplying both sides by 2.

$\theta = 120^\circ + 360^\circ k$

Now we let k = 0, 1, 2, 3, ..., and we see which values will produce an answer in the interval [0, 360).

okdutchman7 · Answer

so 120 is the only value between 0 and 360

okdutchman7 · Answer

480 is too big

mathstudent55 · Answer

$\theta = 120^\circ + 360^\circ k$

For k = 0, 

$\theta = 120^\circ + 360^\circ \times 0 = 120^\circ$

For k = 1,

$\theta = 120^\circ + 360^\circ \times 1 = 120^\circ + 360^\circ = 480^\circ$

which is not in the interval [0, 360).

The only answer is $\theta = 120^\circ$

mathstudent55 · Answer

Correct.

okdutchman7 · Answer

Thanks!

mathstudent55 · Answer

You're welcome.