If 2 cos 2A = ${\sqrt3}$ and A is acute, find :

(i) A

(ii) sin 3A

(iii) sin² (75° - A) + cos² (45° + A)

(i) 2 cos 2A = ${\sqrt3}$

⇒ cos 2A = $\dfrac{\sqrt3}{2}$

⇒ cos 2A = cos 30°

So, 2A = 30°

⇒ A = $\dfrac{30°}{2} = 15°$

Hence, A = 15°.

(ii) sin 3A

= sin (3 x 15°)

= sin 45°

= $\dfrac{1}{\sqrt2}$

Hence, sin 3A = $\dfrac{1}{\sqrt2}$.

(iii) sin² (75° - A) + cos² (45° + A)

= sin² (75° - 15°) + cos² (45° + 15°)

= sin² 60° + cos² 60°

$= \Big(\dfrac{\sqrt3}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{3}{4} + \dfrac{1}{4}\\[1em] = \dfrac{3 + 1}{4} \\[1em] = \dfrac{4}{4} \\[1em] = 1$

Hence, sin² (75° - A) + cos² (45° + A) = 1.