Trig identity to reduce:$\sin^2(u)\cos^2(v)+\sin^2(u)\sin^2(v)+\cos^2(u)\cos^2(v)+\cos^2(u)\sin^2(v)\ $

Question

I am part way through solving the jacobian of a parameterized function. I'm at the point where if i can get rid of the following trig functions I will have a simple function remaining.

Is there anything I can do with this to reduce it down to $~1~$?

$$\sin^2(u)\cos^2(v)+\sin^2(u)\sin^2(v)+\cos^2(u)\cos^2(v)+\cos^2(u)\sin^2(v)\ $$ or $$\sin^2(u)\sin^2(v)+\cos^2(u)\cos^2(v)+\cos^2(u)\sin^2(v)\ $$ Thanks

Answer 1

Note that $$[\sin (u+v)]^2=(\sin u\cos v+ \sin v\cos u)^2=\sin^2 u\cos^2 v +\sin^2 v\cos^2 u +2\sin u\sin v\cos u\cos v$$ and $$[\cos (u+v)]^2=(\cos u\cos v- \sin u\sin v)^2=\cos^2 u\cos^2 v +\sin^2 u\sin^2 v -2\sin u\sin v\cos u\cos v$$

So we have $$\sin^2 u\cos^2 v + \sin^2 v\cos^2 u + \cos^2 u\cos^2 v +\sin^2 u\sin^2 v=\sin^2(u+v)+\cos^2(u+v)=1$$

Also $$\sin^2(u)\sin^2(v)+\cos^2(u)\cos^2(v)+\cos^2(u)\sin^2(v)\ =1-\sin^2u\cos^2v=(1-\sin u\cos v)(1+\sin u\cos v)$$

Answer 2

$\sin^2\left(u\right)\cos^2\left(v\right) + \sin^2\left(u\right)\sin^2\left(v\right) + \cos^2\left(u\right)\cos^2\left(v\right) + \cos^2\left(u\right)\sin^2\left(v\right)$ =

$\left(\cos^2\left(u\right)+\sin^2\left(u\right)\right)\left(\cos^2\left(v\right)+\sin^2\left(v\right)\right)$ =

1

Answer 3

Based on Samuel Bodansky's answer, here is a visual proof:

Now, $\left(\cos^2\left(u\right)+\sin^2\left(u\right)\right)\left(\cos^2\left(v\right)+\sin^2\left(v\right)\right) = 1$ because of the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$.