Question

1/2[sin(2θ + 7θ) + sin(2θ - 7θ)] = _____

cos2θcos7θ
cos2θsin7θ
sin2θcos7θ
sin2θsin7θ

Answer 1

= 1/2[sin2Acos7A + cos2Asin7A + sin2Acos7A - cos2Asin7A]

we cut out

cos2Asin7A - cos2Asin7A

then we have

=1/2[sin2Acos7A + sin2Acos7A ]

= 1/2*2[ sin2Acos7A ]

cut out 2 we get

#sin2Acos7A