GRE Study Precalc

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	\((-1)^{n}\frac{a_{0}}{a_{n}} \)

Resource summary

	Created by Marissa Miller over 8 years ago