How do you express $cos( (5 pi)/6 ) * cos (( 23 pi) /12 )$ without using products of trigonometric functions?

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Answer 1 · 2016-04-21T16:15:07

$=1/2 cos((33pi)/12)+1/2 cos((13pi)/12)$

Use Property:
$cos(A+B)+cos(A-B)=2cosAcosB$

$1/2 [cos(A+B)+cos(A-B)]=cosAcosB$

$A=(5pi)/6, B=(23pi)/12$

$cos((5pi)/6)cos((23pi)/12)=1/2 [cos((5pi)/6+(23pi)/12)+cos((5pi)/6-(23pi)/12)]$

$=1/2 [cos((33pi)/12)+cos((-13pi)/12)]$

$=1/2 cos((33pi)/12)+1/2 cos((13pi)/12)$

How do you express cos( (5 pi)/6 ) * cos (( 23 pi) /12 ) cos( (5 pi)/6 ) * cos (( 23 pi) /12 ) without using products of trigonometric functions?