How do you express $cos (\frac{4 π}{3}) \cdot cos (\frac{π}{6})$ $cos( (4 pi)/3 ) * cos (( pi) /6 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{4 π}{3}) \cdot cos (\frac{π}{6})$ $cos( (4 pi)/3 ) * cos (( pi) /6 )$ without using products of trigonometric functions?

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Answer 1 · 2016-04-01T03:17:24

$- \frac{\sqrt{3}}{2}$ $- sqrt3/2$

Explanation:

$P = cos (\frac{4 π}{3}) . cos (\frac{π}{6})$ $P = cos ((4pi)/3).cos (pi/6)$
Trig table --> $cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ $cos (pi/6) = sqrt3/2$
$cos (\frac{4 π}{3}) = cos (\frac{π}{3} + \frac{3 π}{3}) = cos (\frac{π}{3} + π) = - cos (\frac{π}{3}) = - \frac{1}{2}$ $cos ((4pi)/3) = cos (pi/3 + (3pi)/3) = cos (pi/3 + pi) = - cos (pi/3) = -1/2$
$P = (- \frac{1}{2}) (\frac{\sqrt{3}}{2}) = - \frac{\sqrt{3}}{4}$ $P = (-1/2)(sqrt3/2) = - sqrt3/4$

Answer 2 · 2016-04-04T17:46:30

$cos (\frac{4 π}{3}) cos (\frac{π}{6}) = \frac{1}{2} cos (\frac{9 π}{6}) + \frac{1}{2} cos (\frac{7 π}{6}) = - \frac{\sqrt{3}}{4}$ $cos ((4pi)/3) cos (pi/6)=1/2 cos ((9pi)/6) +1/2 cos ((7pi)/6)=-sqrt3/4$

Explanation:

$2 cos A cos B = cos (A + B) + cos (A - B)$ $2 cos A cos B=cos (A+B) + cos (A-B)$
$cos A cos B = \frac{1}{2} (cos (A + B) + cos (A - B))$ $cos A cos B=1/2 (cos (A+B) + cos (A-B))$
$A = \frac{4 π}{3}, B = \frac{π}{6}$ $A=(4pi)/3, B = pi/6$
$cos (\frac{4 π}{3}) cos (\frac{π}{6}) = \frac{1}{2} (cos (\frac{4 π}{3} + \frac{π}{6}) + cos (\frac{4 π}{3} - \frac{π}{6}))$ $cos ((4pi)/3) cos (pi/6)=1/2 (cos ((4pi)/3+pi/6) + cos ((4pi)/3-pi/6))$
$= \frac{1}{2} (cos (\frac{9 π}{6}) + cos (\frac{7 π}{6}))$ $=1/2 (cos ((9pi)/6) + cos ((7pi)/6))$
$= \frac{1}{2} cos (\frac{9 π}{6}) + \frac{1}{2} cos (\frac{7 π}{6})$ $=1/2 cos ((9pi)/6) +1/2 cos ((7pi)/6)$
$= \frac{1}{2} (0) + \frac{1}{2} (- \frac{\sqrt{3}}{2})$ $=1/2 (0) + 1/2 (-sqrt3/2)$
$= - \frac{\sqrt{3}}{4}$ $=-sqrt3/4$
$cos (\frac{4 π}{3}) cos (\frac{π}{6}) = \frac{1}{2} cos (\frac{9 π}{6}) + \frac{1}{2} cos (\frac{7 π}{6}) = - \frac{\sqrt{3}}{4}$ $cos ((4pi)/3) cos (pi/6)=1/2 cos ((9pi)/6) +1/2 cos ((7pi)/6)=-sqrt3/4$

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How do you express $cos (\frac{4 π}{3}) \cdot cos (\frac{π}{6})$ $cos( (4 pi)/3 ) * cos (( pi) /6 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{4 π}{3}) \cdot cos (\frac{π}{6})$ $cos( (4 pi)/3 ) * cos (( pi) /6 )$ without using products of trigonometric functions?