Trigonometry Challenge

Since we have $\cos \left( 6 \cdot \cos^{-1}\left(\frac{1}{3} \right) \right)$ , let's take $\cos^{-1} \left( \frac{1}{3} \right)=\theta$ $\Rightarrow \cos\theta=\frac{1}{3}$ .

Now we have to find $\cos6\theta$ .

From Triple Angle Identities , we can derive:

$\cos6\theta \ =4\cdot(\cos2\theta)^3-3\cos2\theta$

$\qquad \quad =4\cdot(2\cdot\cos^2\theta-1)^3-3\cdot(2\cos^2\theta-1)$

$\qquad \quad =4\cdot \left(2 \times \frac{1}{9}-1\right)^3-3 \left(2 \times \frac{1}{9}-1\right)$

$\qquad \quad=\dfrac{-1372}{729}+\dfrac{21}{9}$

$\qquad \quad =\dfrac{329}{729}$

$\Rightarrow B-A = 729-329=400. \square$