# Physics equations/01-Introduction/A:mathReview

### Common misconceptions

${\displaystyle \left({\frac {1}{x}}+{\frac {1}{y}}\right)^{-1}\neq x+y}$   and   ${\displaystyle {\sqrt {a^{2}+b^{2}}}\neq a+b}$.

### Percent

The ${\displaystyle X\%}$ symbol means ${\displaystyle X/100}$. A quick and dirty way to find the percent difference is to divide the big number by the small:

${\displaystyle {\frac {BIG}{SMALL}}=1+\underbrace {\frac {BIG-SMALL}{SMALL}} _{percent\;difference}}$

### Trigonometry

In this right triangle: sin A = a/c; cos A = b/c; tan A = a/b.

${\displaystyle \sin A={\frac {\textrm {opposite}}{\textrm {hypotenuse}}}={\frac {a}{\,c\,}}\,.}$   ${\displaystyle \cos A={\frac {\textrm {adjacent}}{\textrm {hypotenuse}}}={\frac {b}{\,c\,}}\,.}$   ${\displaystyle \tan A={\frac {\textrm {opposite}}{\textrm {adjacent}}}={\frac {a}{\,b\,}}={\frac {\sin A}{\cos A}}\,.}$

### Logarithms and exponents are inverse functions

${\displaystyle y=b^{x}\iff x=\log _{b}y}$

The ${\displaystyle \iff }$ implies that the statements are equivalent.

The three most common bases are ${\displaystyle b=2,e,10}$.

The natural log is defined as ${\displaystyle \ln y\equiv \log _{e}y}$.

If ${\displaystyle f=f(x)}$ and ${\displaystyle g=g(y)}$ are inverse functions, then:

${\displaystyle g(f(x))=x}$ and ${\displaystyle f(g(y))=y}$, and we write:

${\displaystyle f=g^{-1}}$ and ${\displaystyle g=f^{-1}}$.

${\displaystyle f^{-1}\neq {\frac {1}{f}}}$.

Complexities occur when the inverse is not a true function, or equivalently, when the inverse is multi-valued:

${\displaystyle \tan ^{-1}(\tan \theta )=\theta \;or\;\theta +\pi }$

Here the problem arises because,

${\displaystyle \tan(\theta )=\tan(\theta +\pi )}$,

so that knowing the tangent of angle does not precisely tell you what the angle was.

${\displaystyle \tan ^{-1}}$ is called the 'arctangent', or the 'inverse tangent'. ${\displaystyle \sin ^{-1}}$ is called 'arcsine', or the 'inverse sine' and so forth.

This quadratic equation, ${\displaystyle ax^{2}+bx+c=0}$, has the solutions:
${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}},}$
If ${\displaystyle f(x)\cdot g(x)\cdot h(x)=0}$ then ${\displaystyle f(x)=0\;or\;g(x)=0\;or\;h(x)=0}$
If ${\displaystyle x(x-2)(x-5)=0}$ then ${\displaystyle x=0\;or\;x=2\;or\;x=5}$