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

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 ?

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

• 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.)
– anon
Dec 1 '13 at 1:15
• Note that the right-hand-side can be reduced further to just $2 \sin \alpha \cos \alpha$. Dec 1 '13 at 1:16
• To explain what @tylerc0816 is saying: $$\sin^2 \alpha + \cos^2 \alpha = 1.$$ Dec 1 '13 at 1:16
• Actually it can be reduced to sin(2a) ;) Dec 1 '13 at 1:17
• Lesson to learn from this: Staring at an equation is not always a good way to understand it. Dec 1 '13 at 1:18

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.
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.