Question

express sin5A-sin3A as a product

Answer 1

sin5A - sin3A = 2 cos[(5A+3A)/2]sin[(5A-3A)/2]        

[ using sinC - sin D = 2 cos[(C+D)/2]sin[(C-D)/2 ]

So sin5A - sin3A = 2 cos(8A/2)sin(2A/2)

                           = 2 cos(4A) sin (A)

⇒ sin5A - sin3A = 2 cos4A sinA

Answer 2

Answer: 2 cos4A sinA

Step-by-step explanation:

We have to Express $\sin 5A -\sin 3A$ as a product

Now as we know

$sinx- siny = 2cos (\dfrac{x+y}{2} )sin(\dfrac{x-y}{2} )$

By formula

$\sin 5A -\sin 3A = 2 \cos (\dfrac{5A+3A}{2} ) \sin (\dfrac{5A-3A}{2} )\\\\= 2\cos (\dfrac{8A}{2} )\sin (\dfrac{2A}{2} )= 2 \cos 4A \sin A$

Hence, sin5A - sin3A is 2 cos4A sinA as product