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While I was solving trigonometry, I came across this peculiar answer from the book. $$ \frac{\sqrt{2+\sqrt3}}{2} $$

Another answer I calculated however was

$$ \frac{\sqrt2+\sqrt6}{4} $$

and both of these values are numerically the same $\approx 0.965926$.

My question is

Can you simplify the former expression to appear as the latter one?

Any help would be appreciated..

N. F. Taussig
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Philippe Duke of Orléans
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4 Answers4

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You could maybe use the fact that both are real positive numbers and their squares are the same. Therefore they have to be the same (otherwise they would be opposites of each other, so one would be negative).

The square of the first is $$\frac{2+\sqrt{3}}{4}$$ and for the second we get $$\frac{2+6+2\sqrt{12}}{16}=\frac{8+4\sqrt{3}}{16}$$

ashsan98
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This is a standard problem in denesting square roots.

Render

$\sqrt{2+\sqrt3}=\sqrt{x}+\sqrt{y}$

which comes with a conjugate equation

$\sqrt{2-\sqrt3}=\sqrt{x}-\sqrt{y}$

Multiply these equations together and note that $2^2-3$, which appears on the left side under the outer square root, is a perfect square:

$1=x-y$

Now square both of the first two equations and add. All the radicals cancel and you can also divide out a factor of $2$ from the sum to get

$2=x+y$

Conclude from the two derived linear equations that $x=3/2,y=1/2$, so

$\sqrt{2+\sqrt3}=\sqrt{3/2}+\sqrt{1/2}.$

Proceed from there.

Oscar Lanzi
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Observe that \begin{align*} \frac{\sqrt{2 + \sqrt{3}}}{2} & = \frac{2\sqrt{2 + \sqrt{3}}}{4}\\ & = \frac{\sqrt{4(2 + \sqrt{3})}}{4}\\ & = \frac{\sqrt{8 + 4\sqrt{3}}}{4}\\ & = \frac{\sqrt{2 + 4\sqrt{3} + 6}}{4}\\ & = \frac{\sqrt{2 + 2 \cdot 2\sqrt{3} + 6}}{4}\\ & = \frac{\sqrt{2 + 2\sqrt{12} + 6}}{4}\\ & = \frac{\sqrt{(\sqrt{2} + \sqrt{6})^2}}{4}\\ & = \frac{\sqrt{2} + \sqrt{6}}{4} \end{align*}

N. F. Taussig
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You need to change the denominator from $2$ to $4$, so multiply be $\frac{2}{2}$:

$$\frac{2\sqrt{2+\sqrt{3}}}{4} = \frac{ \sqrt{2}\sqrt{2} \sqrt{2+\sqrt{3}}}{4}.$$

We split the $2$ in to $\sqrt{2}\sqrt{2}$ because we're aiming to get a $\sqrt{6}$ eventually. Push one $\sqrt{2}$ in side the last radical:

$$= \frac{\sqrt{2}\sqrt{4+2\sqrt{3}}}{4}$$

Our only hope is if the last radicand is a perfect square:

$$4+2\sqrt{3} = 1+2\sqrt{3} + 3 = (1+\sqrt{3})^2$$

So the above is

$$= \frac{\sqrt{2}(1+\sqrt{3})}{4} = \frac{\sqrt{2}+\sqrt{6}}{4}.$$

B. Goddard
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