How does one prove such an equation?

Question

The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.

for $b$ between $-\frac{1}2$ and $1$ :

$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$

Answer 1

Hint:

set $b=\cos t$:

\begin{align} \sqrt{2+\cos t+\sqrt3\sqrt{1-\cos^2t}} &\overset?= \sqrt2\sqrt{\vphantom{\sqrt3}1-\cos t}+ \sqrt{2+\cos t-\sqrt3\sqrt{1-\cos^2t}} ,\\ \sqrt{2+\cos t+\sqrt3\sin t} &\overset?= \sqrt2\sqrt{\vphantom{\sqrt3}1-\cos t}+ \sqrt{2+\cos t-\sqrt3\sin t} ,\\ \sqrt{1+\tfrac12\cos t+\tfrac{\sqrt3}2\sin t} &\overset?= \sqrt{\vphantom{\tfrac{\sqrt3}2}1-\cos t}+ \sqrt{1+\tfrac12\cos t-\tfrac{\sqrt3}2\sin t} \\ \dots \\ \text{(use trigonometry to combine and get rid of radicals)} \\ \dots, \end{align}

and it turns out that on the given interval both sides are equivalent to the same expression \begin{align} \sin(\tfrac12t+\tfrac\pi3) . \end{align}

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Edit:

\begin{align} \sqrt{1+\cos\tfrac\pi3\cos t+\sin\tfrac\pi3\sin t} &\overset?= \sqrt{\vphantom{\tfrac{\sqrt3}2}1-(1-2\sin^2\tfrac t2)}+ \sqrt{1+\cos\tfrac\pi3\cos t-\sin\tfrac\pi3\sin t} ,\\ \sqrt{1+\cos(t-\tfrac\pi3)} &\overset?= \sqrt{\vphantom{\tfrac{\sqrt3}2}1-(1-2\sin^2\tfrac t2)}+ \sqrt{1+\cos(t+\tfrac\pi3)} ,\\ \sqrt{1+2\cos^2\frac{t-\tfrac\pi3}2-1} &\overset?= \sqrt2\sin\tfrac t2+ \sqrt{1+2\cos^2\frac{t+\tfrac\pi3}2-1} ,\\ \cos(\tfrac t2-\tfrac\pi6) &\overset?= \sin\tfrac t2+ \cos(\tfrac t2+\tfrac\pi6) ,\\ \cos\tfrac t2\cos\tfrac\pi6+ \sin\tfrac t2\sin\tfrac\pi6 &\overset?= \sin\tfrac t2+ \cos\tfrac t2\cos\tfrac\pi6- \sin\tfrac t2\sin\tfrac\pi6 ,\\ \cos\tfrac t2\cos\tfrac\pi6+ \sin\tfrac t2\sin\tfrac\pi6 &\overset?= \cos\tfrac t2\cos\tfrac\pi6+ \sin\tfrac t2\sin\tfrac\pi6 . \end{align}

Answer 2

Hint:

Square

$$\sqrt{2+\sqrt{3-3b^2}+b}- \sqrt{2-\sqrt{3-3b^2}+b} = \sqrt{2-2b}.$$

This gives

$$2+\sqrt{3-3b^2}+b+2-\sqrt{3-3b^2}+b-2\sqrt{(2+\sqrt{3-3b^2}+b)(2-\sqrt{3-3b^2}+b)}= 2-2b,$$

$$4+2b-2\sqrt{(2+b)^2-(3-3b^2)}= 2-2b,$$

$$2+4b-2\sqrt{4b^2+4b+1}= 0.$$

Remains to discuss the domain and the signs.

Answer 3

First, we rearrange the equation into the equivalent equation $$ \sqrt{2+b+\sqrt{3-3b^2}}-\sqrt{2+b-\sqrt{3-3b^2}} \overset?= \sqrt{2-2b} $$ which we want to prove. Both sides are positive (keep in mind we have not yet proven the equality), so we can square both sides, and equivalence still holds: $$ 4+2b-2\sqrt{(2+b)^2-(3-3b^2)} = \left(\sqrt{2+b+\sqrt{3-3b^2}}-\sqrt{2+b-\sqrt{3-3b^2}}\right)^2 \overset?= 2-2b $$ Ie, $x^2=y^2$ implies $x=y$ if we know $x,y\ge 0$, otherwise the squares could be equal, but the two sides have opposite signs.

Now, simplify the left-hand side, $$ 2-2b = 4+2b-2\sqrt{1+4b+4b^2}= 4+2b-2\sqrt{(2+b)^2-(3-3b^2)} \overset?= 2-2b, $$ and we see that there actually is equality.

Note that I have used throughout the assumption that the expression inside square roots are non-negative numbers. That's where limitations on $b$ come from.