This is the derivation
sin(x+y)=sinxcosy+cosxsiny 1st equation
sin(x−y)=sinxcosy−cosxsiny 2nd equation
subtract 2nd from the 1st
sin(x+y)−sin(x−y)=2⋅cosxsiny
it follows
cosxsiny=12[sin(x+y)−sin(x−y)]
Now, let x=π3 and y=7π8
cos(π3)sin(7π8)=12[sin(π3+7π8)−sin(π3−7π8)]
cos(π3)sin(7π8)=12[sin(8π+21π24)−sin(8π−21π24)]
cos(π3)sin(7π8)=12[sin(29π24)−sin(−13π24)]
cos(π3)sin(7π8)=12[sin(π+5π24)−sin(−13π24)]
Note: sin(−13π24)=−sin(13π24)
so that
cos(π3)sin(7π8)=12[sin(π+5π24)−(−sin(13π24))]
cos(π3)sin(7π8)=12[sin(π+5π24)+sin(13π24)]
cos(π3)sin(7π8)=
12[sinπcos(5π24)+cosπsin(5π24)+sin(13π24)]
cos(π3)sin(7π8)=12[sin(13π24)−sin(5π24)]
final answer
12⋅sin(13π24)−12⋅sin(5π24)
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