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I'm working through Khan Academy's Calculus courses and I came across a step in a problem that simplified the following using differences of squares.

I can't figure out how the difference of squares was found as it doesn't appear to match any other examples of difference of squares that I found online.

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For clarity the transformation performed is:

\begin{equation} 1-2\sin^2 {\theta} = \left(1+\sqrt{2}\sin{\theta}\right)\left(1-\sqrt{2}\sin{\theta}\right) \end{equation}

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  • $\begingroup$ you mean that $(a-b)(a+b)=a^2-b^2$ ? Thats just that with sin function and $\sqrt 2$ ..... $\endgroup$ – Isham Oct 16 '17 at 2:00
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    $\begingroup$ Put $a=1$ and $b=\sqrt{2}\sin\theta$ in the above identity. $\endgroup$ – VJunior Oct 16 '17 at 2:01
  • $\begingroup$ @EpsilonNeighborhoodWatch someone has edited the post we cant see it yet $\endgroup$ – Isham Oct 16 '17 at 2:04
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    $\begingroup$ @Isham Thanks for informing me. I was hoping the entire equation would be put in the question so I've edited it in myself. $\endgroup$ – Sriotchilism O'Zaic Oct 16 '17 at 2:34
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All "Difference of Squares" really is is the use of the equation $(a-b)(a+b)=a^2-b^2$. In the case shown, $a=1$ and $b=\sqrt2\sin\theta$.

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