Simple problem: Alternate between arithmetical forms

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RussianTank

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How could I alternate between forms to show that:

√[2 + √(3)] / 2 = ( √2 + √2√3 ) / 4


The original problem was finding the exact value of cos(15).
The answer to the left I got using the half-angle formula, and answer to the right is through subtraction formula. Both are correct.

But how to alternate between these forms arithmetically?

If for example I start to convert the left to right by multiplying numerator and denominator by 2, I get:

2[√( 2 + √(3) )] / (2x2)

But now I am stuck. I need help from here.


Regards
 
Last edited:
RussianTank said:
How could I alternate between forms to show that:

√[2 + √(3)] / 2 = ( √2 + √2√3 ) / 4


The original problem was finding the exact value of cos(15).
The answer to the left I got using the half-angle formula, and answer to the right is through subtraction formula. Both are correct.

But how to alternate between these forms arithmetically?

If for example I start to convert the left to right by multiplying numerator and denominator by 2, I get:

2[√( 2 + √(3) )] / (2x2)

But now I am stuck. I need help from here. Regards
Click to expand...

It is simple! Use:

√(8 + 4*√3) = √(2 + 6 + 2 * √2 * √6)

continue.....

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
RussianTank said:
How could I alternate between forms to show that:

√[2 + √(3)] / 2 = ( √2 + √2√3 ) / 4


The original problem was finding the exact value of cos(15).
The answer to the left I got using the half-angle formula, and answer to the right is through subtraction formula. Both are correct.

But how to alternate between these forms arithmetically?

If for example I start to convert the left to right by multiplying numerator and denominator by 2, I get:

2[√( 2 + √(3) )] / (2x2)

But now I am stuck. I need help from here.


Regards
Click to expand...
[MATH]\text {IF } \dfrac{\sqrt{2 + \sqrt{3}}}{2} = \dfrac{\sqrt{2} + \sqrt{2} * \sqrt {3}}{4}, \text {THEN}[/MATH]
[MATH]\dfrac{2 + \sqrt{3}}{4} = \dfrac{2 + 2 * \sqrt{2} * ( \sqrt{2} * \sqrt{3}) + 2 * 3}{16} \\ \text {because squares of equals are equal.}[/MATH]
[MATH]\therefore \dfrac{8 + 4\sqrt{3}}{16} = \dfrac{2 + 6 + 2 * 2 \sqrt{3}}{16} = \dfrac{8 + 4\sqrt{3}}{16}. [/MATH]
That conclusion is TRUE. Now can we reverse that logic? If so, we have a proof by what I recollect was called the method of Proclus.

[MATH]\dfrac{8 + 4\sqrt{3}}{16} = \dfrac{8 + 4\sqrt{3}}{16} \implies[/MATH]
[MATH]\dfrac{2 + \sqrt{3}}{4} = \dfrac{2 + (2 * 2 * \sqrt{3}) + 6}{16} \implies[/MATH]
[MATH]\dfrac{2 + \sqrt{3}}{4} = \dfrac{(\sqrt{2})^2 + 2(\sqrt{2})(\sqrt{2}\sqrt{3}) + (\sqrt{2}\sqrt{3})^2}{16} \implies[/MATH]
[MATH]\dfrac{2 + \sqrt{3}}{2^2} = \dfrac{(\sqrt{2} + \sqrt{2}\sqrt{3})^2}{4^2} \implies[/MATH]
[MATH]\sqrt{\dfrac{2 + \sqrt{3}}{2^2}} = \sqrt{\dfrac{(\sqrt{2} + \sqrt{2}\sqrt{3})^2}{4^2}} \implies[/MATH]
[MATH]\dfrac{\sqrt{2 + \sqrt{3}}}{2} = \dfrac{\sqrt{2} + \sqrt{2}\sqrt{3}}{4}\ \text {Q.E.D}[/MATH]
Follow what we did. We assumed that what we are trying to prove is true. Worked from there to find something known to be true. Then we see if we can reverse that logical chain. If we can we have found our proof. It doesn’t always work, but it’s a place to start if you can’t see a direct path.
 
[MATH] \text {From } \dfrac{\sqrt{2 + \sqrt{3}}}{2} \text{ To } \dfrac{\sqrt{2} + \sqrt{2}\sqrt{3}}{4} . [/MATH]

[MATH]\dfrac{\sqrt{2 + \sqrt{3}}}{2} = \dfrac{2 \sqrt{2 + \sqrt{3}}}{4} = \dfrac{ \sqrt{4(2 + \sqrt{3})}}{4} = \dfrac{\sqrt{8 + 4\sqrt{3}}}{4} = \dfrac{\sqrt{6 + 4\sqrt{3} +2}}{4}.[/MATH]
I got to this part, but I was still stuck. I did not really understand what the intent was with splitting 8 to 6 + 2.

I didn't realize that I was looking at a binomial square in its expanded form! So I cheated. I used a simplifier on the net, from which one can take the following steps to bring it to factored form.

1. Replace 6, with 2*3

[MATH] \dfrac{\sqrt{(2)(3) + 4\sqrt{3} +2}}{4}.[/MATH]
2. The rewrite the first and last terms with square root .


[MATH] \dfrac{\sqrt{\sqrt{(2)(3)}^{2} + 4\sqrt{3} +\sqrt{2}^{2}}}{4}.[/MATH]
The binomial square is the following:

[MATH] (a+b)^{2} = a^{2} + 2ab + b^{2}[/MATH]
So we have

[MATH] a^{2} = \sqrt{(2)(3)}^{2} , \text{ } a = \sqrt{(2)(3)} [/MATH]
We have
[MATH] b^{2} = \sqrt{2}^{2} , \text{ } b = \sqrt{2} [/MATH]
Now we have to show that 2ab is in fact

[MATH] 4\sqrt{3} [/MATH]
Which it is since: [MATH] 4\sqrt{3} = (2)(2)\sqrt{3} = 2(\sqrt{2}\sqrt{2})\sqrt{3} [/MATH]
So let's zoom out again. We now have:

[MATH] \dfrac{\sqrt{\sqrt{(2)(3)}^{2} + 4\sqrt{3} +\sqrt{2}^{2}}}{4} = \dfrac{\sqrt{\sqrt{(2)(3)}^{2} + 2\sqrt{2}\sqrt{2}\sqrt{3} +\sqrt{2}^{2}}}{4} .[/MATH]
With the binomial square in mind we now simplify it to:
[MATH] \dfrac{\sqrt{(\sqrt{(2)(3)} +\sqrt{2})^{2}}}{4} [/MATH]
Now take out the outer square

[MATH] \dfrac{\sqrt{(2)(3)} +\sqrt{2})}{4} [/MATH]
Which is

[MATH] \dfrac{\sqrt{2} + \sqrt{2}\sqrt{3}}{4} [/MATH]
Q.E.D.
 
Good work.

But if you do not understand a step one of us makes, you are allowed to ask questions.
 
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