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Assume the number in sin, cos are in degrees.

Prove 
$\frac{\sqrt{2 + \sqrt2}} {2} (-1 + \sqrt 2) = \frac{\sqrt{2 - \sqrt2}} {2}$

Part 1
1. $\sin(45^\circ) = \cos(45^\circ)$ because $\sin(x) = \cos(90^\circ - x)$

Part 2
1. $\sin(2a) = 2\sin(a)\cos(a)$ double angle
2. $\sin(45) = 2\sin(22.5)\cos(22.5)$

Part 3
1. $\cos(2a) = \cos^2(a) - \sin^2(a)$ double angle
2. $\cos(45) = \cos^2(22.5) - \sin^2(22.5)$

Part 4
1. $2\sin(22.5)\cos(22.5) = \cos^2(22.5) - \sin^2(22.5)$ stitch part 2, 3 using part 1
2. $0 = \cos^2(22.5) - 2\sin(22.5)\cos(22.5) - \sin^2(22.5)$
3. let $j = \cos(22.5), k = \sin(22.5)$
4. $0 = j^{2} - 2jk - k^{2}$

Part 5
1. using quadratic formula, you can get j in terms of k and k in terms of j
2. $j = k (1 \pm \sqrt 2)$
3. $k = j (-1 \pm \sqrt 2)$
4. lets just focus on k in terms of j for now
5. $\sin(22.5) = \cos(22.5) (-1 \pm \sqrt 2)$
6. $\sin(22.5) = \cos(22.5) (-1 + \sqrt 2)$ because this is positive the other one is negative. It doesn't make sense $\sin(22.5)$ to have negative value

Part 6

1. $\cos(22.5) = \frac{\sqrt{2 + \sqrt2}} {2}$ By Wikipedia
2. $\sin(22.5) = \frac{\sqrt{2 + \sqrt2}} {2} (-1 + \sqrt 2)$ Substitution using Part 5 Step 6 and Part 6 Step 1
3. $\sin(22.5) = \frac{\sqrt{2 - \sqrt2}} {2}$ By Wikipedia
4. $\sin(22.5) = \frac{\sqrt{2 + \sqrt2}} {2} (-1 + \sqrt 2) = \frac{\sqrt{2 - \sqrt2}} {2}$

Question is

I know they are equal.

$\frac{\sqrt{2 + \sqrt2}} {2} (-1 + \sqrt 2) = \frac{\sqrt{2 - \sqrt2}} {2}$

But how come I can't directly derive from

$\frac{\sqrt{2 + \sqrt2}} {2} (-1 + \sqrt 2)$

$\frac{\sqrt{2 - \sqrt2}} {2}$

using algebra?

They are the same number just in a different form, yet you can't use algebra to equate those two. You have to take a detour to figure out they are equal.

Thu, 06 Feb 2025 02:23 GMT