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I'm staring at it for hours and I can't make it up, can someone tell me why the bit before the = sign is the same as the bit after ?

$$(\sin\alpha+\cos\alpha-1)(\sin\alpha+\cos\alpha+1)=\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha-1$$

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    $\begingroup$ Have you tried actually distributing it on paper instead of just staring? (It would be even quicker if you have square-of-sum and difference-of-squares formulas memorized.) $\endgroup$
    – anon
    Dec 1 '13 at 1:15
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    $\begingroup$ Note that the right-hand-side can be reduced further to just $2 \sin \alpha \cos \alpha$. $\endgroup$
    – tylerc0816
    Dec 1 '13 at 1:16
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    $\begingroup$ To explain what @tylerc0816 is saying: $$ \sin^2 \alpha + \cos^2 \alpha = 1. $$ $\endgroup$ Dec 1 '13 at 1:16
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    $\begingroup$ Actually it can be reduced to sin(2a) ;) $\endgroup$
    – Mazzy
    Dec 1 '13 at 1:17
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    $\begingroup$ Lesson to learn from this: Staring at an equation is not always a good way to understand it. $\endgroup$
    – Carsten S
    Dec 1 '13 at 1:18
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It uses the Difference of Squares property which states that $$ (a+b)(a-b) = a^2 - b^2. $$In this case, let $a=\sin \alpha + \cos \alpha$ and $b=1$, and see what you get. I won't do the whole thing out for you since you haven't shown us an effort.

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Let $u=\sin\alpha+\cos\alpha$; then the lefthand side is $(u-1)(u+1)$, which is simply $u^2-1$. Now multiply out $u^2$, and you’ll see it.

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  • $\begingroup$ 4 seconds this time! Upvote for you. Hope you'll return the favor. :) $\endgroup$ Dec 1 '13 at 1:16

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