# The value of cos

Question:

The value of $\cos ^{2} 6^{\circ}-\cos ^{2} 24^{\circ}$ is

Solution:

$\cos ^{2} 6^{\circ}-\cos ^{2} 24^{\circ}$

$=\left(\cos 6^{\circ}-\cos 24^{\circ}\right)\left(\cos 6^{\circ}+\cos 24^{\circ}\right)$

$=2 \sin \left(\frac{6^{\circ}+24^{\circ}}{2}\right) \sin \left(\frac{24^{\circ}-6^{\circ}}{2}\right) 2 \cos \left(\frac{6^{\circ}+24^{\circ}}{2}\right) \cos \left(\frac{6^{\circ}-24^{\circ}}{2}\right)$

$=2 \sin 15^{\circ} \sin 9^{\circ} 2 \cos 15^{\circ} \cos \left(-9^{\circ}\right)$

$\left[\right.$ using identities $\left.\cos A-\cos B=2 \sin \frac{A+B}{2} \sin \frac{B-A}{2} \cos A+\cos B=2 \cos \frac{A-B}{2} \cos \frac{A+B}{2}\right]$

$=\left(2 \sin 15^{\circ} \cos 15^{\circ}\right)\left(2 \sin 9^{\circ} \cos 9^{\circ}\right)$

$=\sin 30^{\circ} \sin 18^{\circ} \quad($ since $\cos (-\theta)=\cos \theta)$

$=\frac{1}{2} \sin 18^{\circ} \quad(\because 2 \sin \theta \cos \theta=\sin 2 \theta)$

$\therefore \cos ^{2} 6^{\circ}-\cos ^{2} 24^{\circ}=\frac{1}{2} \sin 18^{\circ}$