Formulas and identities of basic functions

Logarithmic and

Exponential functions

   
  $ \displaystyle\log_{a} x = \frac{\ln x}{\ln a}$ $ \displaystyle a^x = e^{x\ln a}$
Trigonometric functions    
  $ \displaystyle\sec x = \frac{1}{\cos x}$ $ \displaystyle\csc x = \frac{1}{\sin x}$
  $ \displaystyle\tan x = \frac{\sin x}{\cos x}$ $ \displaystyle\cot x = \frac{\cos x}{\sin x}$
  $ \displaystyle\cos^2 x + \sin^2 x = 1$ $ \displaystyle 1 - \cos^2 x = \sin^2 x$
  $ \displaystyle 1 + \tan^2 x = \sec^2 x$ $ \displaystyle \sec^2 - 1 = \tan^2 x$
  $ \displaystyle \sin 2x = 2 \sin x \cos x$ $ \displaystyle\sin x = 2\sin\frac{x}{2} \cos\frac{x}{2}
= \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}^\dagger$
  $ \displaystyle \cos 2x = \cos^2 x - \sin^2 x$ $ \displaystyle\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}
= \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}^\dagger$
  $ \displaystyle\sin^2 x = \frac{1 - \cos 2x}{2}$ $ \displaystyle\cos^2 x = \frac{1 + \cos 2x}{2}$
  $ \displaystyle A \cos x + B \sin x = \sqrt{A^2 + B^2}\cos(x - \theta)$ where $ A > 0$ and $ \theta = \tan^{-1}(B/A)$.
Hyperbolic functions    
  $ \displaystyle\sinh x = \frac{e^x - e^{-x}}{2}$ $ \displaystyle\sinh^{-1} x = \ln\left\vert x + \sqrt{x^2 + 1}\right\vert$
  $ \displaystyle\cosh x = \frac{e^x + e^{-x}}{2}$ $ \displaystyle\cosh^{-1} x = \ln\left\vert x + \sqrt{x^2 - 1}\right\vert$
  $ \displaystyle\tanh x
= \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$ $ \displaystyle\tanh^{-1} x = \frac{1}{2}\ln\left\vert\frac{1 + x}{1 - x}\right\vert$
  $ \displaystyle\cosh^2 x - \sinh^2 x = 1$  



$ \dagger$ You may find them useful in the following ``half-angle tangent'' substitution:


Substitution   Identities  
       
$ t = \displaystyle\tan\left(\frac{x}{2}\right)$, $ \displaystyle\frac{x}{2} = \tan^{-1} t$; $ \sin x = \displaystyle\frac{2t}{1 + t^2}$, $ \cos x = \displaystyle\frac{1 - t^2}{1 + t^2}$, $ \displaystyle dx = \frac{2}{1 + t^2} dt$.

  Example. $\displaystyle \int \frac{1 + \sin x}{1 + \cos x} dx = \int \frac{(1 + t)^2}{1 ...
... \hspace{0.2in}\fbox{(A.1) $s = 1 + t^2$,\: $\frac{ds}{dt} = 2t$}\hspace{5.0in}$    
  $\displaystyle \hspace{1.2in} = t + \ln \vert s\vert + C = \tan\frac{x}{2} + \ln \left\vert 1 + \tan^2\frac{x}{2}\right\vert + C$    



Department of Mathematics
Last modified: 2005-10-05