How do you prove #(sin 8A cos A -sin 6Acos 3A )/ (cos 2 A cos A-sin 3Asin 4A) = tan 2A#?

1 Answer
P dilip_k
Dec 2, 2017

#LHS=(sin 8A cos A -sin 6Acos 3A )/ (cos 2 A cos A-sin 3Asin 4A) #

#=(2sin 8A cos A -2sin 6Acos 3A )/ (2cos 2 A cos A-2sin 3Asin 4A) #

#=(sin 9A+sin7 A -sin 9A-sin3A )/ (cos 3 A +cos A-cosA+cos 7A) #

#=(sin7 A -sin3A )/ (cos 3 A +cos 7A) #

#=(2cos5 Asin2A )/ (2cos5A cos 2A) #

#= tan 2A=RHS#

Related questions
Impact of this question
7213 views around the world
You can reuse this answer
Creative Commons License