Keep squaring on both sides until the roots are gone
Square both sides:
the left hand side is:
7 - 4 sqrt(3),
and the rigth hand side is:
(2 - sqrt(3))(2-sqrt(3)) = 4 -4 sqrt(3) + 3 = 7 - 4 sqrt(3).
Now we need the tan 1/2 angle formula:
tan(A/2) = sqrt((1-cos(A))/(1+cos(A))
Now put A=30 degrees, then cos(30)=sqrt(3)/2, SO:
tan(15) = sqrt((1-sqrt(3)/2)/(1+sqrt(3)/2)) = sqrt((2-sqrt(3)/(2+sqrt(3))
but:
(2-sqrt(3)/(2+sqrt(3) = (2-sqrt(3))^2/[(2+sqrt(3))(2-sqrt(3))]
.............................= (4 -4 sqrt(3) +3)/(4-3) = 7-sqrt(3)
So:
tan(15) = sqrt(7 -sqrt(3)) = 2 - sqrt(3)
RonL
Hello, DivideBy0!
Suppose we don't know the RHS . . .
Here is a primitive method for finding the square root.Find: .sqrt[7 - 4·sqrt{3}]
. . . . . . . . . _ - . . ._______
Let: .a + b·√3 .= .√7 - 4·√3 . where a and b are integers
. . . . . . . . . . . . . . . . . . . ._ . . . . . ._______
Square both sides: .(a + b√3)² .= .(√7 - 4·√3)²
. . . . . . . . . . . . . . . . . ._ . . . . . . - . - . . ._
and we have: .a² + 2ab√3 + 3b² .= .7 - 4·√3
. . . . . . . . . . . . . . . . . . . _ . . . . . . . ._
Then: .(a² + 3b²) + (2ab)√3 .= .7 - 4·√3
Equate coefficients: . a² + 3b² .= .7 .[1]
. . . . . . . . . . . . . . . . . 2ab . . = -4 .[2]
And solve the system of equations . . .
From [2], we have: .b = -2/a .[3]
Substitute into [1]: .a² + 3(-2/a)² .= .7
. . which simplifies to: .a^4 - 7a² + 12 .= .0
. . which factors: .(a² - 3)(a² - 4) .= .0
. . . . . . . . . . . . . . . . . . . _
. . and has roots: .a .= .±√3, ±2
Since a is an integer, we will use: .a = 2
Substitute into [3]: .b = -2/2 = -1
. . . . . . . . . . . _______ . . . . . . . _
. . Therefore: .√7 - 4·√3 . = . 2 - √3
Hello, DivideBy0!
It was laziness on my part . . . sorry.Why was +2 chosen over -2? How did you know it was the right one to use? .I didn't
Choosing a = +2, we get: b = -1, _
. . . and the square root is: . 2 - √3
Choosing a = -2, we get: b = 1, - -_
. . . and the square root is: .-2 + √3
As you suspected, there are two square roots for any quantity (except 0):
. . . . _______ . . . . . . . ._
. . . √7 - 4·√3 .= .±(2 - √3)