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

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

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

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

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

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

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

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

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

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