How do you express #cos( (15 pi)/ 8 ) * cos (( 5 pi) /8 ) # without using products of trigonometric functions?

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Answer 1 · 2016-04-04T16:08:26

#cos((15pi)/8)cos ((5pi)/8)=1/2 cos((5pi)/2)+1/2 cos((5pi)/4)=-sqrt2/2#

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

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

#A=(15pi)/8, B=(5pi)/8#

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

#=1/2 (cos((20pi)/8)+cos((10pi)/8)) #

#=1/2 cos((5pi)/2)+1/2 cos((5pi)/4) =0+ -sqrt2/2=-sqrt2/2 #

#cos((15pi)/8)cos ((5pi)/8)=1/2 cos((5pi)/2)+1/2 cos((5pi)/4)=-sqrt2/2#