GRE Study Precalc

Marissa Miller
Flashcards by Marissa Miller, updated more than 1 year ago
Marissa Miller
Created by Marissa Miller over 4 years ago
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Flashcards on GRE Study Precalc, created by Marissa Miller on 10/05/2015.
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Question Answer
\(\log_{b} x = y\) means... \( b^{y} = x\)
\( \log_{b} (x_{1}x_{2}) = \) \( \log_{b} (x_{1}) + \log_{b} (x_{2}) \)
\( \log_{b} (x_{1}/x_{2}) = \) \( \log_{b} (x_{1}) - \log_{b} (x_{2}) \)
\(b^{\log_{b} x} = \) \(x\)
\((\log_{a} b)(\log_{b} x) = \) \( \log_{a} x\)
\(\log_{b} x^{n} = \) \(n\log_{b} x \)
Sine/Cosine Identity \(\sin^{2} x + \cos^{2} x = 1\)
Cotangent/Cosecant Identity \(1 + \cot^{2}x = \csc^{2}x \)
Tangent/Secant Identity \(\tan^{2} x + 1 = \sec^{2} x \)
\(\sin (\alpha + \beta) = \) \(\sin\alpha\cos\beta + \cos\alpha\sin\beta \)
\(\sin (\alpha - \beta) = \) \(\sin\alpha\cos\beta - \cos\alpha\sin\beta \)
\(\cos(\alpha + \beta) = \) \(\cos\alpha\cos\beta - \sin\alpha\sin\beta \)
\(\cos(\alpha - \beta) = \) \(\cos\alpha\cos\beta + \sin\alpha\sin\beta \)
\(\tan(\alpha + \beta) = \) \(\frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta} \)
\(\tan(\alpha - \beta) = \) \(\frac{\tan\alpha - \tan\beta}{1+\tan\alpha\tan\beta} \)
\(\sin 2\theta = \) \(2\sin\theta\cos\theta \)
\(\cos 2\theta = \) \(\cos^{2}\theta - \sin^{2}\theta \)
\(\tan 2\theta = \) \(\frac{2\tan\theta}{1-\tan^{2}\theta} \)
\(\sin\frac{\theta}{2} = \) \(\sqrt{\frac{1-\cos\theta}{2}} \)
\(\cos\frac{\theta}{2} = \) \(\sqrt{\frac{1+\cos\theta}{2}} \)
\(\tan\frac{\theta}{2} = \) \(\frac{\sin\theta}{1 + \cos\theta} \)
Sum of Roots of a Polynomial \(-\frac{a_{n-1}}{a_{n}} \)
Product of Roots of a Polynomial \((-1)^{n}\frac{a_{0}}{a_{n}} \)
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