(i) 2 sin 5π/12 sin π/12 = 1/2

On using the formula,

2 sin A sin B = cos (A – B) – cos (A + B)

2 sin 5π/12 sin π/12 = cos (5π/12 – π/12) – cos (5π/12 + π/12)

= cos (4π/12) – cos (6π/12)

= cos (π/3) – cos (π/2)

= cos (180°/3) – cos (180°/2)

= cos 60° – cos 90°

= 1/2 – 0

= 1/2

Thus proved.

(ii) 2 cos 5π/12 cos π/12 = 1/2

On using the formula,

2 cos A cos B = cos (A + B) + cos (A – B)

2 cos 5π/12 cos π/12 = cos (5π/12 + π/12) + cos (5π/12 – π/12)

= cos (6π/12) + cos (4π/12)

= cos (π/2) + cos (π/3)

= cos (180°/2) + cos (180°/3)

= cos 90° + cos 60°

= 0 + 1/2

= 1/2

Thus proved.

(iii) 2 sin 5π/12 cos π/12 = (√3 + 2)/2

On using the formula,

2 sin A cos B = sin (A + B) + sin (A – B)

2 sin 5π/12 cos π/12 = sin (5π/12 + π/12) + sin (5π/12 – π/12)

= sin (6π/12) + sin (4π/12)

= sin (π/2) + sin (π/3)

= sin (180°/2) + sin (180°/3)

= sin 90° + sin 60°

= 1 + √3

= (2 + √3)/2

= (√3 + 2)/2

Thus proved.