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# Pythagoras’ Theorem and other Trigonometric Rules

### Compound angles rules

The compound angles rules are:

• $$\cos(A\pm B) = \cos A \cos B \mp \sin A \sin B$$
• $$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$
• When $$A = B$$ we can use the double angle formulae:
$\sin 2A = 2\sin A \cos A$ and \begin{eqnarray*}
\cos 2A &=& \cos^{2}A - \sin^{2}A \\
&=& 1-2\sin^{2}x \\
&=& 2\cos^{2}x-1
\end{eqnarray*}

For example, re-write $$\sin 3x$$ in terms of $$\sin x$$
\begin{eqnarray*}
\sin 3x &=& \sin(2x + x) \\
&=& \sin2x \cos x + \cos 2x \sin x
\end{eqnarray*}
Now using the addition formula ($$\sin 2x = 2 \sin x\cos x$$) and  the double angle formula ($$\cos 2x = 1-2 \sin^{2}x$$) we get:
\begin{eqnarray*}
&&\sin2x \cos x + \cos 2x \sin x \\
&=& (2\sin x\cos x) \cos x + (1-2\sin^{2}x)\sin x \\
&=& 2\sin x \cos^{2}x + \sin x - 2 \sin^{3}x
\end{eqnarray*}
Using the identity ($$\cos^{2}x + \sin^{2}x = 1$$) we have
\begin{eqnarray*}
&&2\sin x \cos^{2}x + \sin x - 2 \sin^{3}x \\
&=& 2 \sin x(1-\sin^{2}x) +\sin x - 2\sin^{3}x \\
&=& 2\sin x - 2 \sin^{3}x + \sin x - 2\sin^{3}x \\
&=& 3 \sin x - 4 \sin^{3}x
\end{eqnarray*}