Prove $ (r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$ How would I verify the following trig identity?
$$(r\sin A \cos A)^2+(r\sin A \sin A)^2+(r\cos A)^2=r^2$$
My work thus far is
$$(r^2\cos^2A\sin^2A)+(r^2\sin^2A\sin^2A)+(r^2\cos^2A)$$
But how would I continue? My math skills fail me.
 A: Just use the distributive property and $\sin^2(x)+\cos^2(x)=1$:
$$
\begin{align}
&(r\sin(A)\cos(A))^2+(r\sin(A)\sin(A))^2+(r\cos(A))^2\\
&=r^2\sin^2(A)(\cos^2(A)+\sin^2(A))+r^2\cos^2(A)\\
&=r^2\sin^2(A)+r^2\cos^2(A)\\
&=r^2(\sin^2(A)+\cos^2(A))\\
&=r^2\tag{1}
\end{align}
$$
This can be generalized to
$$
\begin{align}
&(r\sin(A)\cos(B))^2+(r\sin(A)\sin(B))^2+(r\cos(A))^2\\
&=r^2\sin^2(A)(\cos^2(B)+\sin^2(B))+r^2\cos^2(A)\\
&=r^2\sin^2(A)+r^2\cos^2(A)\\
&=r^2(\sin^2(A)+\cos^2(A))\\
&=r^2\tag{2}
\end{align}
$$
$(2)$ verifies that spherical coordinates have the specified distance from the origin.
A: Oh I didn't read robjohn's answer carefully before making this colourful answer.. I will leave it here anyways.
To continue on what you have:
$$ (\color{red}{r^2}\cos^2A\sin^2A)+(\color{red}{r^2}\sin^2A\sin^2A)+(\color{red}{r^2}\cos^2A) \\
= \color{red}{r^2}( \cos^2A\color{blue}{\sin^2A}+\sin^2A\color{blue}{\sin^2A}+\cos^2A) \\
= r^2( (\color{red}{\cos^2A+\sin^2A})\color{blue}{\sin^2A}+\cos^2A) \\
= r^2( (\color{red}{1})\sin^2A+\cos^2A) \\
= r^2( \color{red}{\cos^2A+\sin^2A}) \\
= r^2( \color{red}{1} ) \\
= r^2
$$
