(i) Given as sin 5π/18 – cos 4π/9 = √3 sin π/9
Let us consider the LHS
sin 5π/18 – cos 4π/9 = sin 5π/18 – sin (π/2 – 4π/9) (since, cos A = sin (90° – A))
= sin 5π/18 – sin (9π – 8π)/18
= sin 5π/18 – sin π/18
On using the formula,
sin A – sin B = 2 cos (A + B)/2 sin (A - B)/2
= 2 cos (6π/36) sin (4π/36)
= 2 cos π/6 sin π/9
= 2 cos 30° sin π/9
= 2 × √3/2 × sin π/9
= √3 sin π/9
= RHS
Thus proved.
(ii) cos π/12 – sin π/12 = 1/√2
Let us consider the LHS
cos π/12 – sin π/12 = sin (π/2 – π/12) – sin π/12 (since, cos A = sin(90° – A))
= sin (6π – 5π)/12 – sin π/12
= sin 5π/12 – sin π/12
On using the formula,
sin A – sin B = 2 cos (A + B)/2 sin (A - B)/2
= 2 cos (6π/24) sin (4π/24)
= 2 cos π/4 sin π/6
= 2 cos 45° sin 30°
= 2 × 1/√2 × 1/2
= 1/√2
= RHS
Thus proved.
(iii) sin 80° – cos 70° = cos 50°
sin 80° = cos 50° + cos 70°
Therefore, now let us consider the RHS
cos 50° + cos 70°
On using the formula,
cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2
cos 50° + cos 70° = 2 cos (50° + 70°)/2 cos (50° – 70°)/2
= 2 cos 120°/2 cos (-20°)/2
= 2 cos 60° cos (-10°)
= 2 × 1/2 × cos 10° (since, cos (-A) = cos A)
= cos 10°
= cos (90° – 80°)
= sin 80° (since, cos (90° – A) = sin A)
= LHS
Thus proved.
(iv) Given as sin 51° + cos 81° = cos 21°
Let us consider the LHS
sin 51° + cos 81° = sin 51° + sin (90° – 81°)
= sin 51° + sin 9° (since, sin (90° – A) = cos A)
On using the formula,
sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2
sin 51° + sin 9° = 2 sin (51° + 9°)/2 cos (51° – 9°)/2
= 2 sin 60°/2 cos 42°/2
= 2 sin 30° cos 21°
= 2 × 1/2 × cos 21°
= cos 21°
= RHS
Thus proved.
