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Friday, 20 May 2011 00:00

Addition and Double Angle Formulae

By  Gideon
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From the definition of cosine:

\[ \cos{\theta} =\frac{e^{i\theta} + e^{-i\theta}}{2} \]

we find:

\[ 2\cos{(\alpha + \beta)} = e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)} .\]

Completing the square gives:

\[ 2\cos{(\alpha + \beta)} \]

\[ = (e^{i\alpha} + e^{-i\alpha})(e^{i\beta} + e^{-i\beta}) - e^{i(\alpha - \beta)} – e^{-i( \alpha - \beta)} \]

\[ = 4\cos{\alpha} \cos{\beta} – 2 \cos{ (\alpha - \beta)} . \]

Or:

\[ \cos{(\alpha + \beta)} = 2\cos{\alpha} \cos{\beta} – \cos{ (\alpha - \beta)} . \]

Alternatively, we can factor the expression this way:

\[ 2\cos{(\alpha + \beta)} = e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)} \]

\[ = (e^{i\alpha} - e^{-i\alpha})(e^{i\beta} - e^{-i\beta}) + e^{i( \alpha - \beta)} + e^{-i(\alpha - \beta)} \]

\[ = -4 \sin{\alpha} \sin{\beta} + 2 \cos{(\alpha - \beta)} . \]

Or:

\[ \cos{(\alpha + \beta)} = -2\sin{\alpha} \sin{\beta} + \cos{( \alpha - \beta)} . \]

Adding the two expressions for \[ \cos{(\alpha + \beta)} \]

We find:

\[ 2\cos{(\alpha + \beta)} \]

\[ = 2\cos{\alpha} \cos{\beta} – \cos{ (\alpha - \beta)} +2\sin{\alpha} \sin{\beta} + \cos{ (\alpha - \beta)} \]

\[ = 2\cos{\alpha} \cos{\beta} - 2\sin{\alpha} \sin{\beta} ,\]

or:

\[ \cos{(\alpha + \beta)} = \cos{\alpha} \cos{\beta} - \sin{\alpha} \sin{\beta} . \]

Setting \[ \alpha = \beta \]

gives:

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

 

Now look at the formula for sine:

\[ \sin{\theta} = \frac{e^{i\theta} - e^{-i\theta}}{2i} \]

It follows that:

\[ 2i \sin{(\alpha + \beta)} = e^{i(\alpha + \beta)} – e^{-i(\alpha + \beta)} \]

\[ = (e^{i\alpha} + e^{-i\alpha} ) ( e^{i\beta} – e^{-i\beta} ) + e^{i(\alpha - \beta)} - e^{-i(\alpha - \beta)} \]

\[ = 4i \cos{\alpha} \sin{\beta} + 2i \sin{(\alpha - \beta)} . \]

Or:

\[ \sin{(\alpha + \beta)} = 2\cos{\alpha} \sin{\beta} + \sin{(\alpha - \beta)} . \]

Alternatively:

\[ 2i \sin{(\alpha + \beta)} = e^{i(\alpha + \beta)} – e^{-i(\alpha + \beta)} \]

\[ = (e^{i\alpha} - e^{-i\alpha} ) ( e^{i\beta} + e^{-i\beta} ) - e^{i(\alpha - \beta) } + e^{-i(\alpha - \beta)} \]

\[ = 4i \sin{\alpha} \cos{\beta} - 2i \sin{(\alpha - \beta)} . \]

Or:

\[ \sin{(\alpha + \beta)} = 2\sin{\alpha} \cos{\beta} - \sin{(\alpha - \beta)} . \]

Adding the two expressions for \[ \sin{(\alpha + \beta)} \]

we find:

\[ 2\sin{(\alpha + \beta)} \]

\[ = 2\cos{\alpha} \sin{\beta} + \sin{(\alpha - \beta)} + 2\sin{\alpha} \cos{\beta} - \sin{(\alpha - \beta)} \]

\[ = 2\cos{\alpha} \sin{\beta} + 2\sin{\alpha} \cos{\beta} , \]

or:

\[ \sin{(\alpha + \beta)} = \cos{\alpha} \sin{\beta} + \sin{\alpha} \cos{\beta} . \]

Setting \[ \alpha = \beta ,\]

gives:

\[ \sin{(2\alpha)} = 2\cos{\alpha} \sin{\alpha}. \]

Read 2549 times Last modified on Tuesday, 01 April 2014 19:11
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