How do you evaluate $cos (- \frac{π}{6}) cos (\frac{π}{4})$ $cos(-pi/6) cos (pi/4)$ ?

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How do you evaluate $cos (- \frac{π}{6}) cos (\frac{π}{4})$ $cos(-pi/6) cos (pi/4)$ ?

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Answer 1 · 2016-06-17T02:33:43

$\frac{\sqrt{6}}{4}$ $sqrt6/4$

Explanation:

Trig table -->
$cos (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $cos (pi/4) = sqrt2/2$
$cos (- \frac{π}{6}) = cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ $cos (-pi/6) = cos (pi/6) = sqrt3/2$
$P = cos (\frac{π}{4}) . cos (- \frac{π}{6}) = (\frac{\sqrt{2}}{2}) (\frac{\sqrt{3}}{2}) = \frac{\sqrt{6}}{4}$ $P = cos (pi/4).cos (-pi/6) = (sqrt2/2)(sqrt3/2) = sqrt6/4$

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How do you evaluate $cos (- \frac{π}{6}) cos (\frac{π}{4})$ $cos(-pi/6) cos (pi/4)$ ?

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How do you evaluate $cos (- \frac{π}{6}) cos (\frac{π}{4})$ $cos(-pi/6) cos (pi/4)$ ?