How do you find the exact value of #sin(pi/12)sin(pi/4)#?

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Answer 1 · 2016-05-20T06:06:21

#sin(pi/12)sin(pi/4)=(sqrt3-1)/4#

We know that #cos(A+B)=cosAcosB-sinAsinB# ..............(1)

and #cos(A-B)=cosAcosB+sinAsinB# ..............(2)

Now subtracting (2) from (1)

#2sinAsinB=cos(A-B)-cos(A+B)# or

#sinAsinB=1/2cos(A-B)-1/2cos(A+B)#

Hence, #sin(pi/12)sin(pi/4)=1/2cos((pi/12)-(pi/4))-1/2cos((pi/12)+(pi/4))#

= #1/2[cos((pi/12)-((3pi)/12))-cos((pi/12)+((3pi)/12))]#

= #1/2[cos(-(2pi)/12)-cos((4pi)/12)]#

= #1/2[cos(-pi/6)-cos(pi/3)]=1/2[cos(pi/6)-cos(pi/3)]#

= #1/2[sqrt3/2-1/2]=(sqrt3-1)/4#