# MandysNotes

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:

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