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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

$(r^2\cos^2A\sin^2A)+(r^2\sin^2A\sin^2A)+(r^2\cos^2A)$

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

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    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.2012-07-18

2 Answers 2

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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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    Thanks for your response I could not have done that on my own.2012-07-19
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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 $