# Differences of Squares on $1-2\sin^2(x)$

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. 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}

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

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 $b=\sqrt2\sin\theta$.