How do you express sin(pi/ 6 ) * cos( ( 3 pi) / 4 ) sin(π6)cos(3π4) without using products of trigonometric functions?

1 Answer

Using Sum or Difference
1/2*sin ((11pi)/12)-1/2*sin ((7pi)/12)12sin(11π12)12sin(7π12)

Explanation:

From "Sum and Difference Formulas"

sin (x+y)=sin x cos y + cos x sin ysin(x+y)=sinxcosy+cosxsiny
sin (x-y)=sin x cos y - cos x sin ysin(xy)=sinxcosycosxsiny

Add the equations to obtain

sin (x+y)+sin(x-y)=2*sin x cos ysin(x+y)+sin(xy)=2sinxcosy

so that, after dividing both sides by 2

sin x cos y=1/2*sin(x+y)+1/2*sin(x-y)sinxcosy=12sin(x+y)+12sin(xy)

at this point , Let x=pi/6x=π6 and y=(3pi)/4y=3π4

sin x cos y=1/2*sin(x+y)+1/2*sin(x-y)sinxcosy=12sin(x+y)+12sin(xy)

sin (pi/6) cos ((3pi)/4)=1/2*sin(pi/6+(3pi)/4)+1/2*sin(pi/6-(3pi)/4)sin(π6)cos(3π4)=12sin(π6+3π4)+12sin(π63π4)

sin (pi/6) cos ((3pi)/4)=1/2*sin((11pi)/12)+1/2*sin((-7pi)/12)sin(π6)cos(3π4)=12sin(11π12)+12sin(7π12)

but sin((-7pi)/12)=-sin((7pi)/12)sin(7π12)=sin(7π12)

therefore

sin (pi/6) cos ((3pi)/4)=1/2*sin((11pi)/12)-1/2*sin((7pi)/12)sin(π6)cos(3π4)=12sin(11π12)12sin(7π12)

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