How do you express #sin(pi/ 6 ) * cos( ( 3 pi) / 4 ) # without using products of trigonometric functions?

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Answer 1 · 2016-02-16T08:07:02

Using Sum or Difference
#1/2*sin ((11pi)/12)-1/2*sin ((7pi)/12)#

From "Sum and Difference Formulas"

#sin (x+y)=sin x cos y + cos x sin y#
#sin (x-y)=sin x cos y - cos x sin y#

Add the equations to obtain

#sin (x+y)+sin(x-y)=2*sin x cos y#

so that, after dividing both sides by 2

#sin x cos y=1/2*sin(x+y)+1/2*sin(x-y)#

at this point , Let #x=pi/6# and #y=(3pi)/4#

#sin x cos y=1/2*sin(x+y)+1/2*sin(x-y)#

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

#sin (pi/6) cos ((3pi)/4)=1/2*sin((11pi)/12)+1/2*sin((-7pi)/12)#

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

therefore

#sin (pi/6) cos ((3pi)/4)=1/2*sin((11pi)/12)-1/2*sin((7pi)/12)#

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