How do you express #sin(pi/ 4 ) * sin( ( 19 pi) / 12 ) # without using products of trigonometric functions?

2 Answers
Nghi N.
Apr 4, 2016

(sqrt2/2)cos (pi/12)

Explanation:

#P = sin (pi/12).sin ((19pi)/12)#
Trig table --> #sin (pi/4) = sqrt2/2#
Trig unit circle and property of complementary arcs give -->
#sin ((19pi)/12) = sin ((7pi)/12 + pi) = - sin ((7pi)/12) = #
#= - sin (pi/12 + pi/2) = cos (pi/12)#
Finally,
#P = (sqrt2/2)cos (pi/12)#

Bdub
Apr 4, 2016

#sin(pi/4)sin((19pi)/12)=-1/2 cos ((11pi)/6) +1/2 cos ((4pi)/3))=(-sqrt3-1)/4#

Explanation:

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

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

#A=pi/4, B= (19pi)/12#

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

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

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

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

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

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

#sin(pi/4)sin((19pi)/12)=-1/2 cos ((11pi)/6) +1/2 cos ((4pi)/3))=(-sqrt3-1)/4#

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