Simplify #(sin3A-sin5A)/(cos3A+cos5A)#?

1 Answer
Shwetank Mauria
Jan 2, 2018

#(sin3A-sin5A)/(cos3A+cos5A)=-tanA#

Explanation:

#(sin3A-sin5A)/(cos3A+cos5A)#

= #-(sin5A-sin3A)/(cos5A+cos3A)#

= #-(2cos((5A+3A)/2)sin((5A-3A)/2))/(2cos((5A+3A)/2)cos((5A-3A)/2))#

= #-(2cos4AsinA)/(2cos4AcosA)#

= #-tanA#

We have used above identities #sinA-sinB=2cos((A+B)/2)sin((A-B)/2)# and
#cosA+sinB=2cos((A+B)/2)cos((A-B)/2)#

Related questions
Impact of this question
6255 views around the world
You can reuse this answer
Creative Commons License