# Simplifying $\sqrt\frac{\left(a^2\cos^2t+b^2\sin^2t\right)^3}{\left(b^2\cos^2t+a^2\sin^2t\right)^3}$

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

$$\sqrt\frac{\left(a^2\cos^2t+b^2\sin^2t\right)^3}{\left(b^2\cos^2t+a^2\sin^2t\right)^3}$$

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.)

• 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 – Bernstein Jun 28 '19 at 15:36
• 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? – Adrian Keister Jun 28 '19 at 15:43
• this is the quotient of a radius of an ellipse over the radius his rotation by $\pi/2$ – Bernstein Jun 28 '19 at 15:58
• 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 $t$. – Michael Hoppe Jun 28 '19 at 16:00
• yes you are right – Bernstein Jun 28 '19 at 16:01

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.

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 $$\sin^2u+\cos^2 u=1$$ $$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}=\sqrt{\frac{(1-\cos^2 u)\cos^2 t+(1-\sin^2u)\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{(\cos^2 t+\sin^2 t)-(\cos^2 u\cos^2 t+\sin^2u\sin^2t)}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{1}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}-1}$$

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