How do you verify the identity #(cos4x+cos2x)/(sin4x+sin2x)=cot3x#?

Question

4574 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-02T10:13:33

Please see below.

We use the identities #cosA+cosB=2cos((A+B)/2)cos((A-B)/2)#

and #sinA+sinB=2sin((A+B)/2)cos((A-B)/2)#

Hence #cos4x+cos2x=2cos((4x+2x)/2)cos((4x-2x)/2)#

= #2cos3xcosx#

and #sin4x+sin2x=2sin((4x+2x)/2)cos((4x-2x)/2)#

= #2sin3xcosx#

Hence #(cos4x+cos2x)/(sin4x+sin2x)=(2cos3xcosx)/(2sin3xcosx)=cot3x#