Question

B) cos 46°-cos 74°; r) cos 15°+cos 45°;​

Answer 1

$\cos(46) - \cos(74) = - 2 \sin( \frac{46 + 74}{2} ) \times \sin( \frac{46 - 74}{2}) = - 2 \sin(60) \times \sin( - 14) = 2 \sin(60) \times \sin(14)$

$\cos(15) + \cos(45) = 2 \cos( \frac{15 + 45}{2} ) \times \cos( \frac{15 - 45}{2} ) = 2 \cos(30) \times \cos( - 15) = 2 \cos(30) \cos(15)$