Double angle
From the triangle below, \(\tan x = \frac{{\sqrt {13} }}{6}\). Find the exact value of \(\cos 2x\).
Use Pythagoras' theorem to work out the hypotenuse, giving you \(\sin x = \frac{{\sqrt {13} }}{7}\) and \(\cos x = \frac{6}{7}\).
You can use any of the three formulae for \(\cos 2x\).
Using \(\cos 2x = {\cos ^2}x - {\sin ^2}x\)
\(= {\left( {\frac{6}{7}} \right)^2} - {\left( {\frac{{\sqrt {13} }}{7}} \right)^2}\)
\(= \frac{{36}}{{49}} - \frac{{13}}{{49}}\)
\(= \frac{{23}}{{49}}\)