(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.
