How do you write the product #6sin8thetacos3theta# as a sum or difference?

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Answer 1 · 2017-02-08T14:26:37

#6sin8thetacos3theta=3sin11theta+3sin5theta#

As #sin(A+B)=sinAcosB+cosAsinB# and

#sin(A-B)=sinAcosB-cosAsinB#,

adding them and reversing LHS and RHS, we get

#2sinAcosB=sin(A+B)+sin(A-B)#

Hence #6sin8thetacos3theta#

= #3xx2sin8thetacos3theta#

= #3xx[sin(8theta+3theta)+sin(8theta-3theta)]#

= #3(sin11theta+sin5theta)#

= #3sin11theta+3sin5theta#