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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*}