If A = 45°, verify that :

(i) sin 2A = 2 sin A cos A

(ii) cos 2A = (2 cos²A - 1) = (1 - 2 sin²A)

(i) L.H.S. :

sin 2A = sin 2(45°) = sin 90°

= 1

R.H.S. :

2 sin A cos A = 2 sin 45° cos 45°

= $2\times\dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}}$

= 1.

Hence, proved that sin 2A = 2 sin A cos A.

(ii) Substituting value of A = 45° in cos 2A, we get :

cos 2A = cos 2(45°)

= cos 90°

= 0.

Substituting value of A = 45° in 2 cos²A - 1, we get :

⇒ 2 cos²A - 1

= 2 cos²45° - 1

= 2 (cos 45°)² - 1

= $2\Big(\dfrac{1}{\sqrt{2}}\Big)^2$ - 1

= $2 \times \dfrac{1}{2} - 1$

= 1 - 1

= 0.

Substituting value of A = 45° in 1 - 2 sin²A, we get :

⇒ 1 - 2 sin²A

= 1 - 2 sin²45°

= 1 - 2 (sin 45°)²

= 1 - $2\Big(\dfrac{1}{\sqrt{2}}\Big)^2$

= 1 - $2 \times \dfrac{1}{2}$

= 1 - 1

= 0.

Hence, proved that cos 2A = (2 cos²A - 1) = (1 - 2 sin²A).