# Prove trig identity: $(\sin \theta + 1)(\sin \theta − 1) = −\cos^2 θ$

This is my attempt:

$(\sin \theta+1)(\sin \theta-1) = \sin\theta^2 - \sin\theta + \sin\theta - 1$

$= \sin^2\theta - 1$

$= -\cos^2\theta$

Is it correct, and can it be improved? Thanks!

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Looks good. Nice work. – John Habert Feb 3 '14 at 5:13
Pretty much as short as it gets. – user61527 Feb 3 '14 at 5:14
It is absolutely correct. – rah4927 Feb 3 '14 at 5:14
Similarly,you can simplify $(\cos\theta+1)(\cos\theta-1)$ – rah4927 Feb 3 '14 at 5:16
Two things - First: You've written $\sin\theta^2$. This should actually (in the context of the question) be $\sin^2\theta$. Second: It is correct, but the step $sinθ^2−sinθ+sinθ−1$ is not required. In fact, you could just use the identity $(a+b)(a-b)=a^2-b^2$. – Sanath K. Devalapurkar Feb 3 '14 at 5:40

Yes this is correct.You must be knowing that $(x+y)(x-y)=x^2-y^2$. Therefore $(sin\theta+1)(sin\theta-1)=sin^2\theta-1=-cos^2\theta$