#2sin3alphacosalpha#?

1 Answer
Narad T.
Jan 27, 2018

The answer is #=cos(2alpha)-cos(4alpha)#

Explanation:

Apply the formula

#sinAsinB=1/2(cos(A-B)-cos(A+B))#

Here,

#A=3alpha#

#B=alpha#

Therefore,

#2sin(3alpha)sin(alpha)=cos(3alpha-alpha)-cos(3alpha+alpha)#

#=cos(2alpha)-cos(4alpha)#

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