Worked examples of problems on Surds as typically found in CIE IGCSE syllabus 0606 (and also CIE GCE O-Level syllabus 4037).
Using \(\sin 15^\circ = \frac{{\sqrt 2 }}{4}\left( {\sqrt 3 - 1} \right)\) and without using a calculator, find the value of \(\sin \theta\) in the form \(a + b\sqrt 2\), where \(a\) and \(b\) are integers.
Solution:
By Sine rule:
\[\frac{{\sin \theta }}{{4\left( {\sqrt 3 + 1} \right)}} = \frac{{\sin 15^\circ }}{{3\sqrt 2 + 4}}\]
\begin{align*}
\sin \theta &= \sin 15^\circ \left( {\frac{{4\left( {\sqrt 3 + 1} \right)}}{{3\sqrt 2 + 4}}} \right)\\
&= \frac{{\sqrt 2 }}{4}\left( {\sqrt 3 - 1} \right)\left( {\frac{{4\left( {\sqrt 3 + 1} \right)}}{{3\sqrt 2 + 4}}} \right)\\
&= \frac{{\sqrt 2 \left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}{{3\sqrt 2 + 4}}\\
&= \frac{{\sqrt 2 \left( {3 - 1} \right)}}{{3\sqrt 2 + 4}}\\
&= \frac{{2\sqrt 2 }}{{3\sqrt 2 + 4}} \times \frac{{3\sqrt 2 - 4}}{{3\sqrt 2 - 4}}\\
&= \frac{{12 - 8\sqrt 2 }}{{18 - 16}}\\
&= 6 - 4\sqrt 2
\end{align*}
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