How do you express $sin(pi/12) * cos(( 7 pi)/12 )$ without products of trigonometric functions?

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Answer 1 · 2018-07-01T09:08:53

$-1/2+sqrt(3)/4$

$sin(pi/12)=(sqrt(3)-1)/(2sqrt(2))$
$cos((7pi)/12)=-(sqrt(3)-1)/(2sqrt(2))$

$sin(pi/12) * cos(( 7 pi)/12 )=-((sqrt(3)-1)/(2sqrt(2)))^2$

$-((sqrt(3)-1)/(2sqrt(2)))^2=-1/8(sqrt(3)-1)^2=-3/8+sqrt(3)/4-1/8=-1/2+sqrt(3)/4$

Answer 2 · 2018-07-01T09:36:10

$color(indigo)(=> (sqrt3 - 2) / 4$

$sin (pi/12) * cos ((7pi) / 12)$

$=> (1/2) (sin (pi/12 + (7pi)/12) + (sin (pi/12) - ((7pi)/12))$

$=> (1/2) (sin ((2pi)/3) + sin (-(pi/2))$

$=> (1/2) (sin (pi/3) - sin (pi/2))$

$=> (1/2) (sqrt3 /2 - 1)$

$color(indigo)(=> (sqrt3 - 2) / 4$

How do you express sin(pi/12) * cos(( 7 pi)/12 ) sin(pi/12) * cos(( 7 pi)/12 ) without products of trigonometric functions?