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I was messing about with a dot product, trying to simplify an expression, when I came across this equality by graphing. Why would these expressions be equal?

$\cos(2x)+r^2=\sqrt{r^4+\frac{r^2h^2}{2}\cos(2x)+\frac{h^4}{16}}$

I noticed that:

$r^4+\frac{r^2h^2}{2}+\frac{h^4}{16}=(r^2+\frac{h^2}{4})^2$

But the $\cos(2x)$ in there really makes it tough. Also, somehow the $h$ factors out completely? Very strange! Hope somebody can figure it out!

Edit: Oops! I happened to only be looking at the function for situations where r >> h. When this isn't true, the equations becomes obviously different. The square rooted equation seemed so close to simplifying though :(

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For fixed $r,x$, the left hand side is constant, but the right hand side varies monotonically with $h$. So the equality cannot hold. To illustrate the point further, for very large $h$, the RHS is large, $\sim \frac{h^2}4$, whereas the LHS doesn't change.

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By squaring both sides, we see that your claimed equality implies $$ \cos^22x + 2r^2\cos 2x+r^4=r^4+\frac{r^2h^2}2\cos2x+\frac{h^4}{16}$$ or equivalently $$\cos^22x+r^2\left(2-\frac{h^2}2\right)\cos 2x-\frac{h^4}{16}=0.$$ In general, this will be false (just plug in $x=\frac\pi4$ and $h\ne0$).

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Suppose this equation does hold for all $h$. Then for $h=0$, the equation reduces to $\cos(2x)=0$. So the overall equation reduces to $$r^2=\sqrt{r^4+h^4/16}> r^2$$ for general $h$. So your equation doesn't hold.

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