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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


But how would I continue? My math skills fail me.

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Sorry for not being able to type my post in mathjax I just lack the technical skill. – jose Jul 18 '12 at 22:28
Look for common factors in each term of the expression. What do you get when you factor these out? – Code-Guru Jul 18 '12 at 22:30
I think you lost a plus sign in there. – Mike Jul 18 '12 at 22:34
You can't solve for $r$, as $r^2$ is a common factor of your equation. If you assume $r \ne 0$ you can divide out the $r^2$. You can prove that it is an identity. – Ross Millikan Jul 18 '12 at 22:41

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.

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Well done, mean square! – ncmathsadist Jul 18 '12 at 23:37
Thanks for your response I could not have done that on my own. – jose Jul 19 '12 at 0:34

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: $$ (\color{red}{r^2}\cos^2A\sin^2A)+(\color{red}{r^2}\sin^2A\sin^2A)+(\color{red}{r^2}\cos^2A) \\ = \color{red}{r^2}( \cos^2A\color{blue}{\sin^2A}+\sin^2A\color{blue}{\sin^2A}+\cos^2A) \\ = r^2( (\color{red}{\cos^2A+\sin^2A})\color{blue}{\sin^2A}+\cos^2A) \\ = r^2( (\color{red}{1})\sin^2A+\cos^2A) \\ = r^2( \color{red}{\cos^2A+\sin^2A}) \\ = r^2( \color{red}{1} ) \\ = r^2 $$

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