How do you express #cos(pi/ 4 ) * sin( ( pi) / 8 ) # without using products of trigonometric functions?

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Answer 1 · 2016-07-04T07:09:33

#cos(pi/4)sin(pi/8)=1/2sin((3pi)/8)-1/2sin(pi/8)#

As #sin(A+B)=sinAcosB+cosAsinB# and

#sin(A-B)=sinAcosB-cosAsinB#

and subtracting second equation from first equation we get

#sin(A+B)-sin(A-B)=2cosAsinB# or

#cosAsinB=sin(A+B)/2-sin(A-B)/2#

Hence #cos(pi/4)sin(pi/8)#

= #sin(pi/4+pi/8)/2-sin(pi/4-pi/8)/2#

= #1/2sin((3pi)/8)-1/2sin(pi/8)#