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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1Looks good. Nice work. – John Habert Feb 03 '14 at 05:13
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1Pretty much as short as it gets. – Feb 03 '14 at 05:14
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1It is absolutely correct. – rah4927 Feb 03 '14 at 05:14
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Similarly,you can simplify $(\cos\theta+1)(\cos\theta-1)$ – rah4927 Feb 03 '14 at 05:16
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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 $(a+b)(a-b)=a^2-b^2$. – Feb 03 '14 at 05:40
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Perfectly correct. Good job. – Claude Leibovici Feb 03 '14 at 06:00
1 Answers
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Yes this is correct.You must be knowing that $(x+y)(x-y)=x^2-y^2$. Therefore $(sin\theta+1)(sin\theta-1)=sin^2\theta-1=-cos^2\theta$
Kamlesh Patel
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