Question

COS2A - COS3A/
sin2A - Sin3A​

Answer 1

Answer:

We know,

cos A – cos B = - 2 sin [(A+B)/2] sin [(A-B)/2]

sin A – sin B = 2 cos [(A+B)/2] sin [(A-B)/2]

tan A = sin A / cos A

Therefore,  

cos 2A – cos 3A  

= -2 sin [(2A+3A)/2] sin [(2A-3A)/2]  

= -2 sin (5A/2) sin (-A/2)  

= 2 sin (5A/2) sin (A/2) …… (i)

And,

sin 2A – sin 3A  

= 2 cos [(2A+3A)/2] sin [(2A-3A)/2]

= 2 cos (5A/2) sin (-A/2)  

= - 2 cos (5A/2) sin (A/2) …… (ii)

Now, by substituting the values from (i) & (ii), let’s solve the given trigonometric equation,

[cos 2A – cos 3A] / [sin 2A – sin 3A]

= [2sin (5A/2) sin (A/2)] / [- 2 cos (5A/2) sin (A/2)]

= sin (5A/2) / - cos (5A/2) [cancelling the similar terms]

= - tan (5A/2)