If A = 45°, verify that:
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = 2 cos2A —1 = 1— 2 sin2A
(i) To show: sin 2A = 2 sin A cos A
A = 45°
∴ To show: sin 90° = 2 sin 45° cos 45°
Consider R.H.S. = 2 sin 45° cos 45°
= 2 × (1/√2) × (1/√2)
= (2 × 1/2)
= 1 = sin 90° = L.H.S.
∴ L.H.S. = R.H.S.
Hence, verified.
(ii) To show: cos 2A = 2 cos2A —1 = 1— 2 sin2A
A = 45°
∴ To show: cos 90° = 2 cos2 45° —1 = 1— 2 sin2 45°
Consider 2 cos2 45° —1 = 1— 2 sin2 45°
= 2 × (1/√2) – 1
= 1 – 1
= 0
= cos 90° = R.H.S.
Consider 1— 2 sin2 45° = 1 – 2(1/2)
= 1 – 1
= 0
= cos 90° = R.H.S.
∴ L.H.S. = R.H.S.
Hence, verified.