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I thought the the Pythagorean identity "$sin^2+cos^2 = 1$" was derived inside the unit circle when the hypotenuse of the triangle was one. So why does this formula work outside of the unit circle? Does the calculator just always assumes that the hypotenuse is one? Would it work if the hypotenuse wasn't one?

Thanks in advance.

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    For a calculator there is probably no hypotenuse and no circle, $\sin(x)$ and $\cos(x)$ are just functions, and $\sin^2(x)+\cos^2(x)=1$ holds as a consequence of $e^{ix}\cdot e^{-ix} = (\cos(x)+i\sin(x))(\cos(x)-i \sin(x))=\cos^2(x)+\sin^2(x)=1$ for any $x\in\mathbb{C}$. – 2017-01-28
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    You multiply by the hypotenuse to scale up unit circle results. Technically I would say that $\sin^2+\cos^2=1$ is related to points *on* the unit circle, not inside it., and when the hypotenuse is less than 1, you scale that way too. – 2017-01-28

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Every triangle outside the unit circle is basically the same as one on the unit circle.

enter image description here

i.e., we can always scale it down and $\sin^2+\cos^2=1$ holds all the way.

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    Hm, why is there a smudge on my triange!?!?! – 2017-01-28
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    Oh, it's not on my screen? I'd better scrape the whiteout off then. – 2017-01-28
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    @SimplyBeautifulArt: that is an essential component of art. Given your nickname, you should leave it there. :D – 2017-01-28
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    Rofl, ok then @JackD'Aurizio – 2017-01-28
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Remember sine and cosine are ratios. Ratos of what?

$$ \sin.. = a/c,\quad \cos.. = b/c $$

Ratio of sides scaled by similar triangles,retaining similarity, to biggest side hypotenuse $1$ instead of to $c.$