Question

prove tan3A= (cos(2A)-cos(4A))/ (sin(4A)- Sin(2A))

Answer 1

cos2A - cos4A = -2 sin(6A/2).sin(-2A/2) = +2 sin(3A).sinA
and
sin4A - sin2A = 2 cos(6A/2).sin(2A/2) = 2 cos(3A).sinA

Hence RHS = (cos(2A)-cos(4A))/ (sin(4A)- Sin(2A)) = sin 3A / cos 3A = tan 3A = LHS

Answer 2

The key identities I used here are

cos x - cos y = -2 sin ((x+y)/2).sin((x-y)/2)
sin x - sin y = 2 cos((x+y)/2).sin((x-y)/2)

Answer 3

Thanks. Is there any other way of solving the equation without the use of those key identities?

Answer 4

I'm sure there are lots of way to deal this. let's try this:
Multiply both sides by sin 4A - sin 2A. Then the original identity is true if

tan 3A ( sin 4A - sin 2A) = cos 2A - cos 4A

which is true if and only if

sin 3A ( sin 4A - sin 2A) = cos 3A ( cos 2A - cos 4A)


Consider the LHS = sin 3A (sin 4A - sin 2A)

Now sin x . sin y = (cos(x-y)-cos(x+y))/2

He ce sin 3A.sin4A - sin 3A.sin2A = 1/2 [ ( cos A - cos 7A) - (cos A - cos 5A) ]
= 1/2 [ cos 5A - cos 7A ]

Now show that the RHS = cos 3A ( cos 2A - cos 4A) equals the same thing using
an expression fo cos x . cos y.