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.


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