# Thread: Formulas Involving Nested Radicals - Trig: (sqrt(2) - sqrt(6))/4 = (-2sqrt(2-sqrt(3))

1. ## Formulas Involving Nested Radicals - Trig: (sqrt(2) - sqrt(6))/4 = (-2sqrt(2-sqrt(3))

Hello all,

Currently, we are covering the trigonometric identities (half-angle, sum, difference, etc.). One particular problem asked to solve for the sin(195deg). So, I first attempted with the sum formula, splitting the 195 into 150 and 45. This yielded the result (sqrt(2) - sqrt(6))/4. However, I then wondered what the result would be utilizing the half-angle formula. Using this, I found the answer to be (-2sqrt(2-sqrt(3))/4. I then set these answers equal to each other and found that -2sqrt(2-sqrt(3)) = sqrt(2) - sqrt(6). Therefore, is there a formula involving nested radicals (like the one found in the half-angle result) that says, for example, what sqrt(A - sqrt(B)) is equivalent to? I am unsure of the correlation between the two numbers on either side of the equation, so any starter help is greatly appreciated.

Thank you!

2. In this particular case we can prove the equality:
sqrt(2) - sqrt(6) = -1*(sqrt(6) - sqrt(2)) = - sqrt( (sqrt(6) - sqrt(2))2 ) = - sqrt( 6 - 2*sqrt(6)*sqrt(2) + 2 )​=
= - sqrt( 8 - 2*sqrt(12) )​ = - sqrt( 8 - 4*sqrt(3) )​= - sqrt( 4(2 - sqrt(3)) ) = -2 sqrt( 2 - sqrt(3) )

I am not aware of a general case.

3. Originally Posted by austint
Hello all,

Currently, we are covering the trigonometric identities (half-angle, sum, difference, etc.). One particular problem asked to solve for the sin(195deg). So, I first attempted with the sum formula, splitting the 195 into 150 and 45. This yielded the result (sqrt(2) - sqrt(6))/4. However, I then wondered what the result would be utilizing the half-angle formula. Using this, I found the answer to be (-2sqrt(2-sqrt(3))/4. I then set these answers equal to each other and found that -2sqrt(2-sqrt(3)) = sqrt(2) - sqrt(6). Therefore, is there a formula involving nested radicals (like the one found in the half-angle result) that says, for example, what sqrt(A - sqrt(B)) is equivalent to? I am unsure of the correlation between the two numbers on either side of the equation, so any starter help is greatly appreciated.

Thank you!
UPDATE: So, I read up on "de-nesting" of radicals and discovered a pattern which read sqrt(X-Y) = sqrt(X) - sqrt(Y). Therefore, the solution to this question is what would sqrt(X - sqrt(Y)) equal to.

4. Originally Posted by austint
UPDATE: So, I read up on "de-nesting" of radicals and discovered a pattern which read sqrt(X-Y) = sqrt(X) - sqrt(Y). Therefore, the solution to this question is what would sqrt(X - sqrt(Y)) equal to.
Huh? This is not true, in general.

E.g. sqrt(49-25) = sqrt(24) = ~4.9

sqrt(49) - sqrt(25) = 7-5 = 2 != 4.9

5. Originally Posted by j-astron
Huh? This is not true, in general.

E.g. sqrt(49-25) = sqrt(24) = ~4.9

sqrt(49) - sqrt(25) = 7-5 = 2 != 4.9
Yes, you're entirely right here. I mistyped the formula here. This is what I found: If sqrt[a + sqrt(b)] = sqrt(x) + sqrt(y),
where a, b, x, and y are rational expressions, and
a greater than sqrt(b), then
sqrt[a - sqrt(b)] = sqrt(x) - sqrt(y).

However, I'm not sure how to derive this.

6. Originally Posted by austint
Yes, you're entirely right here. I mistyped the formula here. This is what I found: If sqrt[a + sqrt(b)] = sqrt(x) + sqrt(y),
where a, b, x, and y are rational expressions, and
a greater than sqrt(b), then
sqrt[a - sqrt(b)] = sqrt(x) - sqrt(y).

However, I'm not sure how to derive this.
I haven't proved it either, but I think I found a counterexample

a = 16
b = 81
x = 4
y = 9

$\sqrt{a + \sqrt{b}} = \sqrt{16 + \sqrt{81}} = \sqrt{16 + 9} = \sqrt{25} = 5$

$\sqrt{x} + \sqrt{y} = \sqrt{4} + \sqrt{9} = 2+3 = 5$

So the first condition is satisfied. Does the statement hold?

$\sqrt{a - \sqrt{b}} = \sqrt{16 - \sqrt{81}} = \sqrt{16 - 9} = \sqrt{7} \approx 2.646$

$\sqrt{x} - \sqrt{y} = \sqrt{4} - \sqrt{9} = 2-3 = -1$

These are not equal to each other. Were there additional conditions that had to be true for the statement to hold?

7. Originally Posted by j-astron
I haven't proved it either, but I think I found a counterexample

a = 16
b = 81
x = 4
y = 9

$\sqrt{a + \sqrt{b}} = \sqrt{16 + \sqrt{81}} = \sqrt{16 + 9} = \sqrt{25} = 5$

$\sqrt{x} + \sqrt{y} = \sqrt{4} + \sqrt{9} = 2+3 = 5$

So the first condition is satisfied. Does the statement hold?

$\sqrt{a - \sqrt{b}} = \sqrt{16 - \sqrt{81}} = \sqrt{16 - 9} = \sqrt{7} \approx 2.646$

$\sqrt{x} - \sqrt{y} = \sqrt{4} - \sqrt{9} = 2-3 = -1$

These are not equal to each other. Were there additional conditions that had to be true for the statement to hold?
I found the statement in Webster Well's "Advanced Course in Algebra" and included that a, b, x, y are rational expressions and a is greater than sqrt(b). However, I believe that verifying the statement only proves that it is true for this qualifications. I want to find actual values of x and y in relation to a and b. Possibly, I can find more equalities by solving more half-angle and double-angle trig identities and then finding a relation between the numbers?

8. Originally Posted by austint
One particular problem asked to solve for the sin(195deg). So, I first attempted with the sum formula, splitting the 195 into 150 and 45. This yielded the result:

. . . . .$\dfrac{\sqrt{2\,}\, -\, \sqrt{6\,}}{4}$

However, I then wondered what the result would be utilizing the half-angle formula. Using this, I found the answer to be:

. . . . .$\dfrac{-2\, \sqrt{2\, -\, \sqrt{3\,}\,}}{4}$

I then set these answers equal to each other and found that -2sqrt(2-sqrt(3)) = sqrt(2) - sqrt(6). Therefore, is there a formula involving nested radicals (like the one found in the half-angle result) that says, for example, what sqrt(A - sqrt(B)) is equivalent to?
There is no formula, that I'm aware of, but there is a methodology.

First let's note that we only really care about the numerators, since the denominators are the same. Also, it would be nicer to deal with non-negative values, so let's work with these:

. . . . .$\sqrt{6\,}\, -\, \sqrt{2\,}\, \mbox{ and }\, 2\, \sqrt{2\, -\, \sqrt{3\,}\,}$

In particular, we would like to show that the right-hand expression is equal to the left-hand expression. We'll assume that the right-hand expression can be stated in terms of a difference of radicals:

. . . . .$2\, \sqrt{2\, -\, \sqrt{3\,}\,}\, =\, \sqrt{a\,}\, -\, \sqrt{b\,}$

Square both sides:

. . . . .$4\, (2\, -\, \sqrt{3\,})\, =\, a\, -\, 2\, \sqrt{ab\,}\, +\, b$

. . . . .$8\, -\, 4\, \sqrt{3\,}\, =\, (a\, +\, b)\, -\, 2\, \sqrt{ab\,}$

Then:

. . . . .$8\, =\, a\, +\, b$

. . . . .$4\, \sqrt{3\,}\, =\, 2\, \sqrt{ab\,}$

The second equation above gives us:

. . . . .$\sqrt{12\,}\, =\, \sqrt{ab\,}$

. . . . .$12\, =\, ab$

. . . . .$\dfrac{12}{a}\, =\, b$

Plug this into:

. . . . .$8\, =\, a\, +\, b$