How do you express #sin(pi/ 4 ) * sin( ( 15 pi) / 8 ) # without using products of trigonometric functions?

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Answer 1 · 2016-03-10T10:26:19

#sin((5pi)/4)sin((15pi)/8)=1/2(cos((5pi)/8)+cos(pi/8))#

For this one uses the identity

#cos(A+B)=cosAcosB-sinAsinB# and
#cos(A-B)=cosAcosB+sinAsinB#.

Subtracting upper identity from lower, one gets

#cos(A-B)-cos(A+B)=2sinAsinB# or

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

Hence #sin((5pi)/4)sin((15pi)/8)#

= #1/2(cos((15pi)/8-(5pi)/4)-cos((15pi)/8+(5pi)/4))#

= #1/2(cos((5pi)/8)-cos((25pi)/8))#

= #1/2(cos((5pi)/8)-cos(3pi+pi/8))#

= #1/2(cos((5pi)/8)+cos(pi/8))#