# GRE Study Precalc

Flashcards by Marissa Miller, updated more than 1 year ago
 Created by Marissa Miller over 4 years ago
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### Description

Flashcards on GRE Study Precalc, created by Marissa Miller on 10/05/2015.

## Resource summary

 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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