Algebra of arctan identity

[Lefiek]

Dear all,

I've attached a picture of an exercise on Khan academy that baffles me.

The angle pi/3 is given with arctan(sqrt(3)) - but following my algebra 9/3sqrt(3) results in 3/sqrt(3).
How do I make the final step - am I missing something obvious?

Thank you for your help

 

[Deleted member 4993]

\(\displaystyle \frac{3}{\sqrt{3}} \ = \ \frac{\sqrt{3}\ \cdot \ \sqrt{3}}{\sqrt{3}} \ = \ \sqrt{3}\)

 

[Lefiek]

Thank you - as I feared and suspected, the answer was obvious.

Side question: how do you visualize the squares like that? It seems like a very useful tool.

 

[Steven G]

By definition sqrt(3) is that special number that when you multiply it by itself gives 3. So sqrt(3)*sqrt(3) = 3 by definition!

 

[Dr.Peterson]

Side question: how do you visualize the squares like that? It seems like a very useful tool.

I think you're asking about the trick of breaking up the 3 in order to cancel one factor. If you don't happen to see that, you can multiply the numerator and denominator by [MATH]\sqrt{3}[/MATH] and then cancel; that is a standard trick called "rationalizing the denominator". Both forms can be very useful.

 

[JeffM]

Thank you - as I feared and suspected, the answer was obvious.

Side question: how do you visualize the squares like that? It seems like a very useful tool.

When you have an expression that is hard to work with, two standard techniques for making life easier are

[MATH]a = a + 0 = a + b - b = (a + b) - b[/MATH]
and [MATH]a = a * 1 = a * \dfrac{b}{b} = \dfrac{ab}{b}\text { with } b \ne 0.[/MATH]
The hard part is seeing what b to use.

What Subhotosh Khan showed you is called rationalizing the denominator. Before hand calculators, it was almost an essential skill.

[MATH]\dfrac{4}{\sqrt{7} - \sqrt{5}} = \dfrac{4}{\sqrt{7} - \sqrt{5}} * 1 = \\ \dfrac{4}{\sqrt{7} - \sqrt{5}} * \dfrac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}} = \dfrac{4(\sqrt{7} + \sqrt{5})}{7 - 5} = 2(\sqrt{7} + \sqrt{5})[/MATH]