Question

If 2 cos 2A =  `sqrt3` and A is acute,
find:
(i) A 
(ii) sin 3A
(iii) sin² (75° - A) + cos² (45° +A)

Answer 1

(i) 2 cos 2A = `sqrt3`

cos 2A = `(sqrt3)/(2)`

cos 2A = cos 30°

2A = 30°

A = 15°

(ii) sin 3A = sin 3(15°)
= sin 45°
= `(1)/(sqrt2)`

(iii) sin²(75° – A ) + cos² (45 + A) = sin² ( 75° –15°) + (cos² ( 45° + 15°) 

= sin² 60° + cos² 60° 

= `(sqrt3/2)^2 + (1/2)^2`

= `(3)/(4) + (1)/(4)`

= 1