Showing $\frac12\sqrt{2+\sqrt2}$ is equivalent to $\frac14\left(\sqrt{4+2\sqrt2} + \sqrt{4-2\sqrt2}\right)$

Question

So I was solving some trigonometry and trying to find value of $\cos 22.5^\circ$ and I used the formula $1+\cos2x=2\cos^2x$ and got the value as $$\frac{\sqrt{2+\sqrt2}}{2}$$ but the answer provided was somewhat like this $$\frac{{\sqrt{4+2\sqrt2} + \sqrt{4-2\sqrt2}}}{4}$$ My book says both are correct and are same representation of same number so how do we convert the first expression the smaller one into the bigger one. I tried to find out by rationalization but was unsuccessful. Please help me out with a way.

Here's a picture of the expressions written in the book Expressions

They both have the same square and they're both positive, so... — J. W. Tanner, Feb 28 '25 at 03:56
As comments above have referred, this is typically about denesting the square/cube roots. we simply use squaring method or conjugate method($(a+\sqrt{b})(a-\sqrt{b})=a^2-b$). — RDK, Feb 28 '25 at 04:33
The question could also have been answered trigonometrically based on its context. If the question had been asked in accordance with this context, it might not have been closed. — , Feb 28 '25 at 05:12

score 0 · Answer 1 · answered Feb 28 '25 at 04:40

When the coefficient of the inner radical is 2, i.e. $\sqrt{a+2\sqrt{d}}$, and wish to write this in the frm of $\sqrt{x}+\sqrt{y}$, you just want two numbers $x>y$ such that

$$x+y=a \text{ and } xy=d$$

For $\sqrt{4+2\sqrt{2}}$, can you find numbers $x+y=4$ and $xy=2$?

Same applies for the $\sqrt{4-2\sqrt{2}}$ being equal to $\sqrt{x}-\sqrt{y}$.

Showing $\frac12\sqrt{2+\sqrt2}$ is equivalent to $\frac14\left(\sqrt{4+2\sqrt2} + \sqrt{4-2\sqrt2}\right)$

1 Answers1