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So there are several trigonometric identities, some very well know, such as $\cos(x) = 1 - 2\sin^2(\frac{x}{2})$ and some more obscure like $\tan(\frac{\theta}{2} + \frac{\pi}{4}) = \sec(\theta)+\tan(\theta)$.

You can also find some logarithmic identities online

But looking for radical identities (stuff involving square roots), if I google I'm lead to politics and ethnicity (???)... I want to find some cool identities involving square roots that are always true for any $x>0$. I am aware they exist, I've seen some, but I don't remember them...

What are some cool obscure identities involving square roots?

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$$\sqrt{1+\sqrt{N}} + \sqrt{N+N\sqrt{N}} - \sqrt{3N+1+(N+3)\sqrt{N}} \ = 0$$


$$\sqrt{ (\sin{\theta}+\csc{\theta})^2 + (\cos{\theta}+\sec{\theta})^2 - (\tan{\theta}-\cot{\theta})^2 } \ = 3$$

Background of the above trigonometric identity:

In the 1896 (but not 1893) version of S. L. Loney’s classic book Plane Trigonometry, I found the following amazing question on page 25:

Prove the following statement.

$$ (\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = \tan^2 \alpha + \cot^2 \alpha + 7 $$

I was so amazed by the $7$. Having such a ‘large’ number in a trigonometric identity is rare.

I managed to make a cool variant of the question with an even larger number:

Simplify:

$ (\sin{\theta}+\csc{\theta})^2 + (\cos{\theta}+\sec{\theta})^2 - (\tan{\theta}-\cot{\theta})^2 $

Answer: $9$

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  • $\begingroup$ The first identity, $\sqrt{3N+1+(N+3)\sqrt{N}}=\sqrt{1+\sqrt{N}}+\sqrt{N+N\sqrt{N}}$ can be unraveled with $(1+M)^{3/2}=\sqrt{1+3M+3M^2+M^3}$ or $=\sqrt{1+M}+M\sqrt{1+M}$, ie, expand a fractional power $x^{a/b}$ as a binomial under a root, $(x^a)^{1/b}$, or as a product with the power's integer part, $x^q\cdot x^r$. Then set $M=\sqrt{N}$ and rearrange. $\endgroup$
    – Jam
    Aug 27, 2023 at 10:14
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Not so obscure, but: [with thanks to user Will for spotting typos]

$\sqrt{xy}=\sqrt x\sqrt y$

$ax^2+bx+c=a\left(x-{-b+\sqrt{b^2-4ac}\over2a}\right)\left(x-{-b-\sqrt{b^2-4ac}\over2a}\right)$

$a^4+b^4=(a^2-\sqrt2ab+b^2)(a^2+\sqrt2ab+b^2)$

$a^5-b^5=(a-b)\left(a^2+{1-\sqrt5\over2}ab+b^2\right)\left(a^2+{1+\sqrt5\over2}ab+b^2\right)$

Here are some more, from G. Chrystal, Textbook of Algebra:

$\sqrt{{x+y\over x-y}}+\sqrt{{x-y\over x+y}}={2x\over\sqrt{x^2-y^2}}$

$\sqrt{{a+\sqrt{a^2-b}\over2}}+\sqrt{{a-\sqrt{a^2-b}\over2}}=\sqrt{a+\sqrt b}$

$\left({1+\sqrt{1-x}\over1-\sqrt{1-x}}\right)^{1/2}+\left({1-\sqrt{1-x}\over1+\sqrt{1-x}}\right)^{3/2}={2(2-x)\over x^2}(\sqrt x-\sqrt{x-x^2})$

$\bigl(\sqrt{p^2+1}+\sqrt{p^2-1}\bigr)^{-3}+\bigl(\sqrt{p^2+1}-\sqrt{p^2-1}\bigr)^{-3}=(p^2-(1/2))\sqrt{p^2+1}$

$\sqrt{\sqrt{a^2+\root3\of{a^4b^2}}+\sqrt{b^2+\root3\of{a^2b^4}}}=(a^{2/3}+b^{2/3})^{3/4}$

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1. Golden ratio: $\phi = \frac{1+\sqrt{5}}{2}$ look at the sequences A001622 , A139339

$$ \sqrt{1 + 2 I} = \sqrt{\phi}+ \frac{I}{\sqrt{\phi}}$$

2. Seven $\pi$: $$ 7\pi = {2^{2^2} \tan ^{-1}\left(1+\sqrt{2}+\sqrt{2 \left(2+\sqrt{2}\right)}\right)}$$

3. Cotangent Radicals:

$$\cot \frac{\pi}{2^{2^{2^2}}} = $$

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

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Half-angle formulas $$ \sin \frac{\theta}{2} = \pm \sqrt{\frac{1-\cos\theta}{2}} \\ \cos \frac{\theta}{2} = \pm \sqrt{\frac{1+\cos\theta}{2}} \\ \tan \frac{\theta}{2} = \pm \sqrt{ \frac{1-\cos\theta}{1+\cos\theta}} $$

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Let $x$ be any real number greater than $1$ so that both of the following iterated square roots converge:

$a=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$

$b=\sqrt{x-\sqrt{x-\sqrt{x-...}}}$

Then

$a^2-b^2=(x+a)-(x-b)=a+b$

and therefore

$a-b=\sqrt{x+\sqrt{x+\sqrt{x+...}}}-\sqrt{x-\sqrt{x-\sqrt{x-...}}}=1.$

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Just a sample,

$$\sqrt[3]{\frac1{9}}+\sqrt[3]{-\frac2{9}}+\sqrt[3]{\frac4{9}}=\left(4\sqrt[3]{\frac2{3}}-5\sqrt[3]{\frac1{3}}\right)^{1/8}$$

$$\sqrt[5]{\frac1{25}}+\sqrt[5]{\frac3{25}}+\sqrt[5]{-\frac9{25}}=\left(\sqrt[5]{\frac{32}{5}}-\sqrt[5]{\frac{27}{5}}\right)^{1/3}$$

$$\sqrt[5]{-\frac1{125}}+\sqrt[5]{\frac2{125}}+\sqrt[5]{\frac8{125}}+\sqrt[5]{\frac{16}{125}}=\left(\sqrt[5]{\frac1{5}}+\sqrt[5]{\frac4{5}}\right)^{1/2}$$

and a lot more by Ramanujan. These are just special cases of more general identities.

When it comes to unusual radicals, there was none more prolific than him. Just google "Ramanujan and radicals", and it should keep you busy for a while. Try:

"Ramanujan's Association with Radicals in India" (by Berndt, et al)

"Radicals and units in Ramanujan's work" (by Berndt, et al)

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