How do you express #sin(pi/12) * cos(( 7 pi)/12 ) # without products of trigonometric functions?

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Answer 1 · 2018-07-01T09:08:53

#-1/2+sqrt(3)/4#

#sin(pi/12)=(sqrt(3)-1)/(2sqrt(2))#
#cos((7pi)/12)=-(sqrt(3)-1)/(2sqrt(2))#

#sin(pi/12) * cos(( 7 pi)/12 )=-((sqrt(3)-1)/(2sqrt(2)))^2#

#-((sqrt(3)-1)/(2sqrt(2)))^2=-1/8(sqrt(3)-1)^2=-3/8+sqrt(3)/4-1/8=-1/2+sqrt(3)/4#

Answer 2 · 2018-07-01T09:36:10

#color(indigo)(=> (sqrt3 - 2) / 4#

#sin (pi/12) * cos ((7pi) / 12)#

#=> (1/2) (sin (pi/12 + (7pi)/12) + (sin (pi/12) - ((7pi)/12))#

#=> (1/2) (sin ((2pi)/3) + sin (-(pi/2))#

#=> (1/2) (sin (pi/3) - sin (pi/2))#

#=> (1/2) (sqrt3 /2 - 1)#

#color(indigo)(=> (sqrt3 - 2) / 4#