How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{11 π}{12})$ $cos( (5 pi)/6 ) * cos (( 11 pi) /12 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{11 π}{12})$ $cos( (5 pi)/6 ) * cos (( 11 pi) /12 )$ without using products of trigonometric functions?

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Answer 1 · 2016-09-13T03:21:31

$P = cos (\frac{5 π}{6}) . cos (\frac{11 π}{12})$ $P = cos ((5pi)/6).cos ((11pi)/12)$
Trig table of special arcs, and unit circle -->
$cos (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2}$ $cos ((5pi)/6) = - sqrt3/2$
$cos (\frac{11 π}{12}) = cos ((- \frac{π}{12}) + π) = - cos (- \frac{π}{12}) = - cos (\frac{π}{12})$ $cos ((11pi)/12) = cos ((- pi/12) + pi) = - cos (-pi/12) = - cos (pi/12)$
P can be expressed as:
$P = (\frac{\sqrt{3}}{2}) . cos (\frac{π}{12})$ $P = (sqrt3/2).cos (pi/12)$

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How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{11 π}{12})$ $cos( (5 pi)/6 ) * cos (( 11 pi) /12 )$ without using products of trigonometric functions?

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How do you express cos(5π6)⋅cos(11π12)cos( (5 pi)/6 ) * cos (( 11 pi) /12 ) without using products of trigonometric functions?

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How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{11 π}{12})$ $cos( (5 pi)/6 ) * cos (( 11 pi) /12 )$ without using products of trigonometric functions?