How do you verify the identity #sin(alpha/3)cos(alpha/3)=1/2sin((2alpha)/3)#?

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Answer 1 · 2017-11-10T19:51:17

see below

#sin(alpha/3)cos(alpha/3)#

#=1/2[2sin(alpha/3)cos(alpha/3)]#

#=1/2[sin(alpha/3)cos(alpha/3)+sin(alpha/3)cos(alpha/3)]--(1)#

but

#sin(x+y)=sinxcosy+sinycosx#

#:.(1)rarr1/2[sin(alpha/3+alpha/3)]#

#=1/2sin((2alpha)/3)#

as required