How do you express #sin(pi/ 4 ) * sin( ( 1 9 pi) / 12 ) # without using products of trigonometric functions?

Question

2286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-20T12:19:06

#sin(pi/4)sin((19pi)/12)=1/2cos((4pi)/3)-1/2cos((11pi)/6)#

We know that

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

#cos(A+B)=cosAcosB-sinAsinB#. Subtracting latter from former, we get

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

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

Hence #sin(pi/4)sin((19pi)/12)=sin((19pi)/12)sin(pi/4)#

= #1/2cos((19pi)/12-pi/4)-1/2cos((19pi)/12+pi/4)#

= #1/2cos((19pi)/12-(3pi)/12)-1/2cos((19pi)/12+(3pi)/12)#

= #1/2cos((16pi)/12)-1/2cos((22pi)/12)#

= #1/2cos((4pi)/3)-1/2cos((11pi)/6)#

Answer 2 · 2016-11-06T02:42:07

#=-1/4(sqrt3+1)#

#sin(pi/4)sin((19pi)/12)#

#=sin(pi/4)sin(pi+(7pi)/12)#

#=-sin(pi/4)sin((7pi)/12)#

#=-sin(pi/4)sin(pi/3+pi/4)#

#=-sin(pi/4)[sin(pi/3)cos(pi/4)+cos(pi/3)sin(pi/4))]#

#=-1/sqrt2[sqrt3/2xx1/sqrt2+1/2xx1/sqrt2]#

#=-1/4(sqrt3+1)#