1. ## Equivalence of 2 expressions involving radicals

Hi.

I need help in seeing how the following two expressions are equal.

$\frac{\sqrt{6} + \sqrt{2}}{4}$ and $\frac{\sqrt{2}\sqrt{1 + \frac{\sqrt{3}}{2}}}{2}$

I know they both evaluate to the same thing but I'm having a hard time seeing how they are related.
If someone could, perhaps through algebraic manipulation, show how one is equivalent to the other,
I would be immensely grateful.

I don't think it matters but the first arrived at by evaluating $cos \frac{\pi}{12}$ using a sum of angles
formula while the second is what you get if you decide to use a half angle formula.

2. ## Re: Equivalence of 2 expressions involving radicals

Hint: Starting with $\frac{\sqrt{6}+\sqrt{2}}{4}$ get the denominator the same as the other expression and then factorise $\sqrt{2}$ out of the numerator

3. ## Re: Equivalence of 2 expressions involving radicals

Or work the other way: start from $\frac{\sqrt{2}\sqrt{1+ \frac{\sqrt{3}}{2}}}{4}$, take the $\sqrt{2}$ inside the other integral: $\frac{\sqrt{2(1+ \frac{\sqrt{3}}{2}})}{4}$, then take one of the "2"s in the denominator into it:
$\frac{\sqrt{\frac{2}{4}(1+ \frac{\sqrt{3}}{2}})}{2}= \frac{\sqrt{\frac{1}{2}(1+ \frac{\sqrt{3}}{2}})}{2}$

4. ## Re: Equivalence of 2 expressions involving radicals

Originally Posted by kodx
I need help in seeing how the following two expressions are equal.
$\frac{\sqrt{6} + \sqrt{2}}{4}$ and $\frac{\sqrt{2}\sqrt{1 + \frac{\sqrt{3}}{2}}}{2}$
Enough time has passed for me to give you a solution. I agree with reply #2.

Note that ${\left( {\frac{{\sqrt 6 + \sqrt 2 }}{4}} \right)^2} = \frac{1}{2} + \frac{{\sqrt 3 }}{4}$, therefore $\frac{{\sqrt 6 + \sqrt 2 }}{4} = \sqrt {\frac{1}{2} + \frac{{\sqrt 3 }}{4}}$

Thus
\begin{align*} \frac{{\sqrt 6 + \sqrt 2 }}{4} &= \sqrt {\frac{1}{2} + \frac{{\sqrt 3 }}{4}}\\ &= \frac{1}{{\sqrt 2 }}\left( {\sqrt {1 + \frac{{\sqrt 3 }}{2}} } \right)\\ &= \frac{{\sqrt 2 \left( {\sqrt {1 + \frac{{\sqrt 3 }}{2}} } \right)}}{2}\end{align*}