We know that cos(A+B)=cosAcosB-sinAsinBcos(A+B)=cosAcosB−sinAsinB ..............(1)
and cos(A-B)=cosAcosB+sinAsinBcos(A−B)=cosAcosB+sinAsinB ..............(2)
Now subtracting (2) from (1)
2sinAsinB=cos(A-B)-cos(A+B)2sinAsinB=cos(A−B)−cos(A+B) or
sinAsinB=1/2cos(A-B)-1/2cos(A+B)sinAsinB=12cos(A−B)−12cos(A+B)
Hence, sin(pi/4)sin(pi/3)=1/2cos((pi/4)-(pi/3))-1/2cos((pi/4)+(pi/3))sin(π4)sin(π3)=12cos((π4)−(π3))−12cos((π4)+(π3))
= 1/2cos(-pi/12)-1/2cos((7pi)/12)12cos(−π12)−12cos(7π12),
but cos(-x)=cosxcos(−x)=cosx and cos(pi-x)=-cosxcos(π−x)=−cosx,
hence above is equal to
1/2cos(pi/12)-1/2cos(pi-(5pi)/12)12cos(π12)−12cos(π−5π12)
= 1/2cos(pi/12)+1/2cos((5pi)/12)=1/2[cos(pi/12)+cos((5pi)/12)]12cos(π12)+12cos(5π12)=12[cos(π12)+cos(5π12)]