How do you express #sin(pi/12) * cos(( 23 pi)/12 ) # without products of trigonometric functions?

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Answer 1 · 2016-01-15T00:50:24

You can use the sum-to-product identities ...

#(sinA-sinB)/2=cos((A+B)/2)sin((A-B)/2)#

Now, solve for A and B ...

#(A-B)/2=pi/12#

#(A+B)/2=(23pi)/12#

#A=2pi#

#B=(11pi)/6#

Now, insert A and B back into the first part of the identity...

#(sin(2pi)-sin((11pi)/6))/2=0.5#

hope that helped!