## Trigonometry (11th Edition) Clone

$$(\sin x+1)^2-(\sin x-1)^2=4\sin x$$
$$A=(\sin x+1)^2-(\sin x-1)^2$$ We can use the identity: $$a^2-b^2=(a-b)(a+b)$$ That means $$A=[(\sin x+1)-(\sin x-1)][(\sin x+1)+(\sin x-1)]$$ $$A=[(\sin x-\sin x)+(1-(-1))][(\sin x+\sin x)+(1-1)]$$ $$A=[0+2][2\sin x+0]$$ $$A=2\times2\sin x$$ $$A=4\sin x$$