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Angle sum and difference identities

It can be shown that

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$
$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$
$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin\beta$
$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin\beta$

Prerequisite information[edit]

The proof[edit]

Euler's formula says that $e^{ix} = \cos x + i \sin x$. For $x = \alpha + \beta$ we have

[math]e^{i(\alpha + \beta)} = \cos(\alpha + \beta) + i \sin(\alpha + \beta)\ \ \ \ \ \ \ \ \ \ (1)[/math]

However, we also know that $e^{x+y} = e^x e^y$. Thus

[math]e^{i(\alpha + \beta)} = e^{i \alpha} e^{i \beta}.[/math]

Using Euler's formula again,

[math]\begin{array} e^{i \alpha} e^{i \beta} &=& (\cos \alpha + i \sin \alpha) (\cos \beta + i \sin \beta) \\ &=& \cos \alpha \cos \beta+ i(\sin \alpha \cos \beta + \cos \alpha \sin \beta) - \sin \alpha \sin\beta \\ &=& (\cos \alpha \cos \beta - \sin \alpha \sin\beta) + i(\sin \alpha \cos \beta + \cos \alpha \sin \beta) \ \ \ \ \ \ \ \ \ \ (2) \end{array}[/math]

Matching real and imaginary parts of (1) and (2), we have

[math]\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin\beta[/math]

[math]\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta[/math]

Replacing $\beta$ by $-\beta$

[math]\cos(\alpha -\beta ) = \cos \alpha \cos \beta + \sin \alpha \sin\beta[/math]

[math]\sin(\alpha -\beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta[/math]