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```
Date: 06/18/2001 at 21:19:03
From: Alec Colvin

I was playing around with radical expression for the trig functions
when I noticed the following:

sin(15) = sqrt(2-sqrt(3))/2 = (sqrt(6)-sqrt(2))/4

How do you manipulate sqrt(2-sqrt(3))/2 to obtain (sqrt(6)-sqrt(2))/4?

Thanks.
```

```
Date: 06/19/2001 at 10:15:07
From: Doctor Peterson

Hi, Alec.

Not all expressions of this form can be simplified; but if one can, we
would like the square root to have the form

sqrt(a) + sqrt(b)  or  sqrt(a) - sqrt(b)

where a and b are positive rational numbers, and a > b to give the
positive square root.

If we assume that sqrt(2-sqrt(3)) has this form, then this must be
true:

(sqrt(a) +- sqrt(b))^2 = 2 - sqrt(3)

Expanding, we get

a + b +- 2 sqrt(ab) = 2 - sqrt(3)

Equating the rational and irrational parts of this equation,

a + b = 2
+-2 sqrt(ab) = -sqrt(3)

The second equation here tells us that we will have to choose the
negative sign in order to have a solution, since square roots have to
be positive. If we now square this equation, we have

a + b = 2
4ab = 3

Solving these equations by substituting in the second, we get

4a(2 - a) = 3

4a^2 - 8a - 3 = 0

8 +- sqrt(64 + 48)
a = ------------------ = 1/2 or 3/2
8

and

b = 2 - a = 3/2 or 1/2.

One choice here will give the positive root, the other the negative
root. We take a > b, so that a = 3/2 and b = 1/2. Consequently, our
solution is

sqrt(2 - sqrt(3)) = sqrt(3/2) - sqrt(1/2) = (sqrt(6) - sqrt(2))/2

Looking for information on this technique, which I didn't recall being
techniques includes a quote from an old algebra text about the method:

Fun with algebraic number theory - Dave Rusin

Here's a page I just found in our archives about another approach:

http://mathforum.org/dr.math/problems/tran.03.16.01.html

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents
High School Polynomials

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