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

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

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

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

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

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