# MandysNotes

13 November 2012

${\color{Goldenrod}\sin}\ \left( {\color{MidnightBlue}z_{1}} + {\color{MidnightBlue}z_{2}} \right) = {\color{Goldenrod}\sin}\ {\color{MidnightBlue}z_{1}} \ {\color{Blue}\cos}\ {\color{MidnightBlue}z_{2}} \ + \ {\color{Blue}\cos}\ {\color{MidnightBlue}z_{1}}\ {\color{Goldenrod}\sin}\ {\color{MidnightBlue}z_{2}} \$

${\color{Blue}\cos}\ \left( {\color{MidnightBlue}z_{1}} + {\color{MidnightBlue}z_{2}} \right) = {\color{Blue}\cos}\ {\color{MidnightBlue}z_{1}} \ {\color{Blue}\cos}\ {\color{MidnightBlue}z_{2}} \ - \ {\color{Goldenrod}\sin}\ {\color{MidnightBlue}z_{1}}\ {\color{Goldenrod}\sin}\ {\color{MidnightBlue}z_{2}} \$

## Derivatives of Trigonometric Functions

13 November 2012

$\frac{d}{dx}{ \color{Goldenrod}\sin}\ {z} = {\color{Blue}\cos}\ {z}$

$\frac{d}{dx}{\color{Blue} \cos}\ {z} = -{\color{Goldenrod} \sin}\ {z}$

## Primitives of Trigonometric Functions

13 November 2012

$\int{ \color{Goldenrod}\sin}\ {z} = - {\color{Blue}\cos}\ {z}$

$\int{\color{Blue} \cos}\ {z} ={\color{Goldenrod} \sin}\ {z}$

## Half Angle Formulae

21 May 2011

We can find the half angle formula for cosine by starting with the double angle formula:

$\cos{(2\alpha)} = \cos^{2}{\alpha} - \sin^{2}{\alpha} \ .$

Making the substitution:

$\gamma = 2\alpha ,$

$\cos{\gamma} = \cos^{2}{(\frac{\gamma}{2})} - \sin^{2}{(\frac{\gamma}{2})} .$