Melody

Are you a clairvoyant Zac?

How did you know what the asker wanted?     

enter as                  asin(-1/2)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)} = -{\mathtt{30^{\circ}}}$$

You need to explain how you get the answer anees :/

Mmm I'll have to think about this one.   I don't think that i get it Nauseated  

I haven't looked at the question either:((

It is a pity that there are just SO many questions.

It is really hard to find the time to really digest some of the great questions and answers that are given :)

Ok, I hate to boost his ego still further but TitaniumRome is right.  

You cannot take the square root to simplify a fraction.  

You have to divide top and botton by the same number and since there are no common factors it cannot be simplified.

You can however change it into a mixed numeral :)

what do you want to convert?

Hi TitaniumRome, 

You have to give explanations - they can probably get the answer only from the back of the book.

Your role as an answerer is to help people learn :)

Hi Nauseated,

Yes it was!!  Want another peice?

Bertie is on now, maybe you would like the icecream Bertie ? 

It is of a no melt variety.  Its makers are very ingenious :)

I don't know - maybe you should take your coat off ?

Now I have to consider

  $$\\\sqrt[3]{-1}\\
$The first obvious answer is $ -1+0i $ \;\;since (-1)^3=-1\\
$Now, there will be 3 roots$\\
$each will be $2\pi/3 $ radians apart. $\\
$So the other 2 will be $\\
=cos(\pi/3)+isin(\pi/3)\qquad and \qquad cos(-\pi/3)+isin(-\pi/3)\\
=\qquad \frac{1}{2}+\frac{\sqrt3i}{2} \qquad \qquad \qquad and\qquad \qquad \frac{1}{2}-\frac{\sqrt3i}{2}\\
=\qquad \frac{1+\sqrt3i}{2} \qquad \qquad \qquad and\qquad\qquad\frac{1-\sqrt3i}{2}$$

NOW

$$\\\sqrt[3]{-1}=-1,\qquad \frac{1+\sqrt3i}{2}, \qquad\frac{1-\sqrt3i}{2}\\
so\\
\sqrt[3]{14\sqrt[3]{-1}}\\\\
=\sqrt[3]{14(-1)},\qquad \sqrt[3]{14*\frac{1+\sqrt3i}{2}}, \qquad \sqrt[3]{14*\frac{1-\sqrt3i}{2}}\\\\
=\sqrt[3]{-14},\qquad \sqrt[3]{7(1+\sqrt3i)}, \qquad \sqrt[3]{7(1-\sqrt3i)}\\\\
$Now WolframAlpha only gave 3 answers but here I am going to get 3 lots of 3 answers, perhaps each lot is the same as the other 3 lots?$\\\\
\sqrt[3]{-14}\\\\
=\sqrt[3]{14}\sqrt[3]{-1}\\\\
=-\sqrt[3]{14},\qquad \sqrt[3]{14}*\frac{1+\sqrt3i}{2}, \qquad\sqrt[3]{14}*\frac{1-\sqrt3i}{2}\\\\$$

$$\\$This approximates to$\\\\
=-2.410,\qquad 2.410*\frac{1+\sqrt3i}{2}, \qquad2.410*\frac{1-\sqrt3i}{2}\\\\
=-2.410,\qquad 2.410*(0.5+\frac{\sqrt3i}{2}), \qquad2.410*(0.5-\frac{\sqrt3i}{2})\\\\
=-2.410,\qquad 1.205+2.087i, \qquad 1.205-2.087i\\\\$$

Badinage is right.   None of these answers agree with Wolfram|alpha