Prove that sin x/cos 3x + sin 3x/cos 9x + sin 9x/cos 27x

 

Question:

Prove that

sin x/cos 3x + sin 3x/cos 9x + sin 9x/cos 27x

= 1/2 (tan 27x - tan x)

Solution:

We know that,

sin A/cos 3A = 1/2 (tan 3A - tan A)

Therefore,

sin x/cos 3x

= 1/2 (tan 3x - tan x)

sin 3x/cos 9x

= 1/2 (tan 9x - tan 3x)

sin 9x/cos 27x

= 1/2 (tan 27x - tan 9x)

Adding,

LHS = 1/2[(tan 3x - tan x)

          + (tan 9x - tan 3x)

          + (tan 27x - tan 9x)]

= 1/2[tan 3x - tan x

      + tan 9x - tan 3x

      + tan 27x - tan 9x]

Middle terms cancel:

= 1/2 (tan 27x - tan x)

Hence proved.

Answer:

1/2 (tan 27x - tan x)