MandysNotes

13 November 2012 In Trigonometric Formulae



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


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

13 November 2012 In Trigonometric Formulae

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

 

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

13 November 2012 In Trigonometric Formulae

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

\[ \int{\color{Blue} \cos}\ {z} ={\color{Goldenrod} \sin}\ {z} \]

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:

\[ \gamma = 2\alpha ,\]

leads to the equation:

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

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