Given that \( \tan 2x = \frac{3}{4} \), \( 0 < x < \frac{\pi}{4} \), find the exact value of (a) \( \cos 2x \) (b) \( \cos x \). - Scottish Highers Maths - Question 10 - 2015

To find ( \cos 2x ), we can use the identity:

$\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$

From the given, ( \tan 2x = \frac{3}{4} ). We can find ( \tan x ) using the relation:

$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$

Let ( \tan x = t ), then:

$\frac{2t}{1 - t^2} = \frac{3}{4}$

Cross-multiplying gives:

$8t = 3(1 - t^2)$

Expanding leads to:

$3t^2 + 8t - 3 = 0$

Using the quadratic formula:

$t = \frac{-4 \pm \sqrt{(4^2) - 3 \cdot 3 \cdot (-3)}}{2 \cdot 3} = \frac{-4 \pm \sqrt{16 + 27}}{6} = \frac{-4 \pm \sqrt{43}}{6}$

Choosing the positive value (since ( 0 < x < \frac{\pi}{4} )) gives:

$\tan x = \frac{-4 + \sqrt{43}}{6}$

Now substituting back into ( \cos 2x ):

$\cos 2x = \frac{1 - \left(\frac{-4 + \sqrt{43}}{6}\right)^2}{1 + \left(\frac{-4 + \sqrt{43}}{6}\right)^2}$

Calculate ( \cos 2x ) to find the exact value.