I am looking to simplify these term [ I forgot the 3 :( ]


where $a$ and $b$ are two non-negative reals.

(This is not homework. I am just trying to make my expression easy, but I didn't find a way.)

Thanks for your help.

  • $\begingroup$ by simple I mean not a quotient, Thanks for all the answers but I have the feeling that it can be just a constant!!!!! I am not sure $\endgroup$
    – Bernstein
    Commented Jun 28, 2019 at 15:36
  • $\begingroup$ I don't think that's possible. You can eliminate either $\sin$ or $\cos,$ but that's about the limit. Perhaps you could provide some larger context for helping us understand what you're trying to do? $\endgroup$ Commented Jun 28, 2019 at 15:43
  • $\begingroup$ this is the quotient of a radius of an ellipse over the radius his rotation by $\pi/2$ $\endgroup$
    – Bernstein
    Commented Jun 28, 2019 at 15:58
  • $\begingroup$ It can't be constant. Geometrically -- for positive $a$ and $b$ -- it's the quotient of the distances from the origin of two ellipses with axis $a$, $b$ and $b$, $a$, resp. in corresponding times $t$. $\endgroup$ Commented Jun 28, 2019 at 16:00
  • $\begingroup$ yes you are right $\endgroup$
    – Bernstein
    Commented Jun 28, 2019 at 16:01

3 Answers 3


Divide both the numerator and denominator by $a^2+b^2$, then use $\frac{a^2}{a^2+b^2}=\sin^2 u$. Your expression will become $$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}$$ Now use $\sin^2u+\cos^2 u=1$ $$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}=\sqrt{\frac{(1-\cos^2 u)\cos^2 t+(1-\sin^2u)\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{(\cos^2 t+\sin^2 t)-(\cos^2 u\cos^2 t+\sin^2u\sin^2t)}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{1}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}-1}$$


The form is $$\sqrt{\frac{f^3}{g^3}}$$ which we can write as $$\left(\frac{f}{g}\right)^{3/2}$$ so let's just worry about that inner quotient, $f/g$.

The quotient is definitely not constant; different $t$ values give different results: $$t = 0 \;\to\; \frac{a^2}{b^2} \qquad\qquad t= \frac{\pi}{2}\;\to\;\frac{b^2}{a^2}$$

Having to set aside this dream case, we're left to consider a few alternatives and decide which might be consider least-bad.

  • Leave it as-is. $$\frac{a^2\cos^2 t + b^2 \sin^2 t}{a^2 \sin^2 t+b^2\cos^2 t} \tag{1}$$ That expression isn't terribly complicated.

  • Divide-through by $\cos^2t$ in the numerator and denominator, to get $$\frac{a^2+b^2\tan^2t}{b^2+a^2\tan^2t} \tag{2}$$ This may not be appreciably better, though ... and it introduces unnecessary concern about $t=\pi/2$.

  • @Dr.SonnhardGraubner's answer invokes the double-angle formula to get something I'll write as $$\frac{\left(a^2+b^2\right)+\left(a^2-b^2\right)\cos 2t}{\left(a^2+b^2\right)-\left(a^2-b^2\right)\cos 2t} \tag{3}$$ which is "simpler" in that the degree of the trig functions is lower, at the cose of adding some complexity to the coefficients.

  • @Andrei's suggestion to trade $a$ and $b$ for trig functions is a good one, although I'd choose to swap the sine and cosine assignment to write $a = \sqrt{a^2+b^2} \cos u$ (and $b = \sqrt{a^2+b^2} \sin u$). I'd also use the double-angle formulas to simplify the resulting quotient to $$\frac{1 + \cos 2t \cos 2u}{1 - \cos 2t \cos 2u} \tag{4}$$ (I'd also probably choose to associate $a$ with $\cos u$ (and $b$ with $\sin u$), which changes the above slightly.) Note that $(3)$ seems to be crying-out for us to make such a substitution.

  • Since the quotient appears in the context of ellipses, we could use $a^2-b^2=c^2$ (where $c$ is the center-to-focus distance) and $c = ea$ (where $e$ is the eccentricity) to write $$\frac{ 1 -e^2 \sin^2 t}{1 -e^2 \cos^2 t} \tag{5}$$ I like this one, personally.

  • One can also combine re-writing in terms of $e$ with re-writing in terms of $\cos 2t$ (left as an exercise to the reader), but this just seems to re-complicate things.

One can imagine other variations, too. How useful any one version is depends upon how it's intended to be used.


For the radicand i have got $$\frac{a^2 (-\cos (2 t))-a^2+b^2 \cos (2 t)-b^2}{a^2 \cos (2 t)-a^2-b^2 \cos (2 t)-b^2}$$


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